Calculating discrete probability - GMAT Quantitative

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Question

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

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Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

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Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

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Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Tap to reveal answer

Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Tap to reveal answer

Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Tap to reveal answer

Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Tap to reveal answer

Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Tap to reveal answer

Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Tap to reveal answer

Answer

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Question

At a school fair, there are 25 water balloons. 10 are yellow, 8 are red, and 7 are green. You try to pop the balloons. Given that you first pop a yellow balloon, what is the probability that the next balloon you hit is also yellow?

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Answer

At the start, there are 25 balloons and 10 of them are yellow. You hit a yellow balloon. Now there are 9 yellow balloons left out of 24 total balloons, so the probability of hitting a yellow next is

$\frac{9}{24}=\frac{3}{8}$ .

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Question

Race cars for a particular race are numbered sequentially from 12 to 115. What is the probability that a car selected at random will have a tens digit of 1?

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Answer

There are 104 integers from 12 to 115 inclusive. There are 8 integers from 12 to 19 and 6 integers from 110 to 115 for a total of 14 integers with a tens digit of 1. The probability of selecting a car with a tens digit of 1 is $\frac{14}{104}$ = $\frac{7}{52}$ .

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Question

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?

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Answer

Prob(new OR yellow) = P(new) + P(yellow) - P(new AND yellow)

Prob(new) = $\dpi{100} \small \frac{45}{100}$

Prob(yellow) = $\dpi{100} \small \frac{25}{100}$

Prob(new AND yellow) = $\dpi{100} \small \frac{45}{100}\times \frac{25}{100}$

so P(new OR yellow) = $\dpi{100} \small \frac{45}{100}+\frac{25}{100}-\frac{45}{100}\times \frac{25}{100}$

$\dpi{100} \small =\frac{70}{100}-\frac{9}{80}$

$\dpi{100} \small =\frac{7}{10}-\frac{9}{80}$

$\dpi{100} \small =\frac{56}{80}-\frac{9}{80}=\frac{47}{80}$

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Question

Flight A is on time for 93% of flights. Flight B is on time for 89% of flights. Flight A and B are both on time 87% of the time. What is the probabiity that at least one flight is on time?

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Answer

P(Flight A is on time) = 0.93

P(Flight B is on time) = 0.89

P(Flights A and B are on time) = 0.87

Then P(A OR B) = P(A) + P(B) - P(A AND B) = 0.93 + 0.89 – 0.87 = 0.95

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Question

A bag has 7 blue balls and 3 red balls. 2 balls are to be drawn successively and without replacement. What is the probability that the first ball is red and the second ball is blue?

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Answer

We first have 7 blue and 3 red out of 10, so P(1st ball is red) = $\frac{3}{10}$ . Now, we have pulled a red ball out of the bag, leaving us with 7 blue and 2 red out of 9 total balls. Then P(2nd ball is blue) = $\frac{7}{9}$ . Put this together, P(1st red AND 2nd blue) = $\frac{3}{10}\cdot \frac{7}{9}=\frac{7}{30}$ .

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Question

4 cards are to be dealt successively and without replacement from an ordinary deck of 52 cards. What is the probability of receiving, in order, a spade, a heart, a diamond, and a club?

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Answer

There are 4 suits, and each suit has 13 cards, so P(spade) = $\frac{13}{52}$ .

Now there are 51 cards left: 12 spades, 13 hearts, 13 diamonds, and 13 clubs, so now P(heart) = $\frac{13}{51}$ .

Once again, there are now 50 cards: 12 spades, 12 hearts, 13 diamonds, and 13 clubs, so now P(diamond) = $\frac{13}{50}$ .

Before our last draw we have 49 cards left: 12 spades, 12 hearts, 12 diamonds, and 13 clubs, so P(club) = $\frac{13}{49}$ .

Putting these four probabilities together gives us our answer:

P(spade AND heart AND diamond AND club) = $\frac{13}{52 }\cdot \frac{13}{51}\cdot \frac{13}{50}\cdot \frac{13}{49}$

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Calculating discrete probability - GMAT Quantitative

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This makes 14 favorable outcomes out of 64 outcomes total, so the probability is .

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This makes 14 favorable outcomes out of 64 outcomes total, so the probability is .

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This makes 14 favorable outcomes out of 64 outcomes total, so the probability is .

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This makes 14 favorable outcomes out of 64 outcomes total, so the probability is .

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This makes 14 favorable outcomes out of 64 outcomes total, so the probability is .

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This makes 14 favorable outcomes out of 64 outcomes total, so the probability is .

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted. What is the probability that the sum will be 7?

At a school fair, there are 25 water balloons. 10 are yellow, 8 are red, and 7 are green. You try to pop the balloons. Given that you first pop a yellow balloon, what is the probability that the next balloon you hit is also yellow?

At the start, there are 25 balloons and 10 of them are yellow. You hit a yellow balloon. Now there are 9 yellow balloons left out of 24 total balloons, so the probability of hitting a yellow next is .

Race cars for a particular race are numbered sequentially from 12 to 115. What is the probability that a car selected at random will have a tens digit of 1?

There are 104 integers from 12 to 115 inclusive. There are 8 integers from 12 to 19 and 6 integers from 110 to 115 for a total of 14 integers with a tens digit of 1. The probability of selecting a car with a tens digit of 1 is = .

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?

Prob(new OR yellow) = P(new) + P(yellow) - P(new AND yellow) Prob(new) = Prob(yellow) = Prob(new AND yellow) = so P(new OR yellow) =

Flight A is on time for 93% of flights. Flight B is on time for 89% of flights. Flight A and B are both on time 87% of the time. What is the probabiity that at least one flight is on time?

P(Flight A is on time) = 0.93 P(Flight B is on time) = 0.89 P(Flights A and B are on time) = 0.87 Then P(A OR B) = P(A) + P(B) - P(A AND B) = 0.93 + 0.89 – 0.87 = 0.95

A bag has 7 blue balls and 3 red balls. 2 balls are to be drawn successively and without replacement. What is the probability that the first ball is red and the second ball is blue?

We first have 7 blue and 3 red out of 10, so P(1st ball is red) = . Now, we have pulled a red ball out of the bag, leaving us with 7 blue and 2 red out of 9 total balls. Then P(2nd ball is blue) = . Put this together, P(1st red AND 2nd blue) = .

4 cards are to be dealt successively and without replacement from an ordinary deck of 52 cards. What is the probability of receiving, in order, a spade, a heart, a diamond, and a club?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

At the start, there are 25 balloons and 10 of them are yellow. You hit a yellow balloon. Now there are 9 yellow balloons left out of 24 total balloons, so the probability of hitting a yellow next is

$\frac{9}{24}=\frac{3}{8}$ .

There are 104 integers from 12 to 115 inclusive. There are 8 integers from 12 to 19 and 6 integers from 110 to 115 for a total of 14 integers with a tens digit of 1. The probability of selecting a car with a tens digit of 1 is $\frac{14}{104}$ = $\frac{7}{52}$ .

P(Flight A is on time) = 0.93

P(Flight B is on time) = 0.89

P(Flights A and B are on time) = 0.87

Then P(A OR B) = P(A) + P(B) - P(A AND B) = 0.93 + 0.89 – 0.87 = 0.95