Calculating discrete probability

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GMAT Quantitative › Calculating discrete probability

Questions 11 - 20

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?

$\frac{7}{32}$

$\frac{1}{4}$

$\frac{7}{64}$

$\frac{1}{8}$

$\frac{5}{18}$

Explanation

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

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

A coin is tossed times, what is the probabilty of having a head showing up on the third toss?

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{32}$

Explanation

This trick question is in fact very simple. Since the throws are independent (in other words, what happened before an event has no influence on the future outcomes), then the probabilty is $\frac{1}{2}$ , indeed there are no reasons for any of the head or tail to have more chance of showing up at any point.

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?

$\frac{3}{8}$

$\frac{2}{5}$

$\frac{9}{25}$

$\frac{5}{12}$

$\frac{3}{5}$

Explanation

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

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

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?

$\frac{7}{52}$

$\frac{14}{103}$

$\frac{1}{13}$

$\frac{8}{103}$

$\frac{13}{104}$

Explanation

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 $\frac{14}{104}$ = $\frac{7}{52}$ .

One hundred marbles - red, yellow, blue, or green - are placed in a box. There are an equal number of red and blue marbles, three times as many yellow marbles as green marbles, and five fewer green marbles than red marbles. What is the probability that a randomly drawn marble is NOT blue?

$\frac{4}{5}$

$\frac{3}{5}$

$\frac{17}{20}$

$\frac{11}{20}$

$\frac{3}{4}$

Explanation

If there are blue marbles, there are also red marbles, green marbles, and $3 \left (x-5 \right )$ yellow marbles. There are 100 marbles total, so we can solve for in the equation:

20 of the 100 marbles are blue, so the probability that the marble is not blue is:

$\frac{100-20}{100} = \frac{80}{100} = \frac{4}{5}$

Target

The upper semicircle of the above target has radius twice that of the lower semicircle.

A blindfolded woman throws a dart at random at the above target. Disregarding any skill factor, and assuming that the dart hits the target, what are the odds against the dart landing inside the purple region?

$\textup{13 to 2}$

$\textup{11 to 1}$

$\textup{7 to 1}$

$\textup{6 to 1}$

$\textup{11 to 2}$

Explanation

For the sake of simplicity, we will assume the smaller semicircle of the target has radius 1, and, consequently, the larger semicircle has radius 2.

The larger semicircle has total area

$A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi$

The smaller semicircle has total area

$A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi$

The total area of the target is

$A = A_{1}+ A_{2} = 2 \pi + \frac{1}{2} \pi = \frac{5}{2} \pi$

The purple sector has area one-sixth that of the larger semicircle, or

$\frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi$

The area that is not purple is

$\frac{5}{2} \pi - \frac{1}{3} \pi = \frac{15}{6} \pi - \frac{2}{6} \pi = \frac{13}{6} \pi$

The odds against the dart landing in the purple region are

$\frac{\frac{13}{6} \pi}{\frac{1}{3} \pi} = \frac{\frac{13}{6} \pi \cdot \frac{6}{\pi}}{\frac{1}{3} \pi \cdot \frac{6}{\pi}} = \frac{13}{2}$ - that is, 13 to 2.

Marianna inserts the joker into a standard deck of 52 cards. By how much has she decreased the probability that a card drawn at random from the deck will be an ace?

$\frac{1}{689}$

$\frac{1}{53}$

$\frac{1}{52}$

$\frac{1}{2,756}$

$\frac{105}{2,756}$

Explanation

The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is an ace (or any other given rank) is $\frac{4}{52}= \frac{1}{13}$ ; if the joker is added, the probability decreases to $\frac{4}{53}$ . The decrease in probability is

$\frac{1}{13} - \frac{4}{53}=\frac{53 \cdot 1}{53 \cdot 13} - \frac{4\cdot 13}{53 \cdot 13} =\frac{53 }{689} - \frac{52}{689} = \frac{1}{689}$

This problem set is designed to have a better understanding of probabilties

We throw two regular six sided dice. What is the probabilty of having a pair?

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{16}$

$\frac{1}{36}$

Explanation

The correct answer is $\frac{1}{6}$ , since there are 2 dice and we are looking for the probabilty than any pair shows up, the first die can then be any number, therefore we assign it a probability of one and the second die must be the same number as the first, and therefore has a probabilty of $\frac{1}{6}$ . Multiplying these two probabilities, we obtain $\frac{1}{6}$ .

Let's try to solve more complicated problems using the same logic as in the dice problems.

We have a standard -card deck. We draw two cards without replacement. What is the probabilty of drawing a pair?

$\frac{3}{51}$

$\frac{1}{26}$

$\frac{1}{52}$

$\frac{1}{104}$

$\frac{1}{2704}$

Explanation

Here we are asked for the probability of drawing any pair. Therefore, the first card can be any card and we can assign this event a probability of one.

Furthermore, the second card drawn must be the same card to form a pair. Therefore, its probability is $\frac{3}{51}$ .

We use 51 because we already have drawn a card. You will notice that to form a pair there are 3 other cards in the deck that can be drawn, for example if an ace is drawn first, there must be three other aces in the deck.

$1\cdot \frac{3}{51}$ is the final answer

We flip a coin times. What is the probabilty of having at least one tail show up?

$\frac{31}{32}$

$\frac{29}{32}$

$\frac{27}{32}$

$\frac{1}{2}$

$\frac{3}{4}$

Explanation

Again since we are asked to look for the probability of an event occuring at least once, we can take a shortcut and calculate the probability of the opposite event; 'no tail shows up'. The probability of this event is given by $\frac{1}{2^{5}}$ or $\frac{1}{32}$ .

We substract this $\frac{1}{32}$ to 1, the sum of all the probabilities, to obtain $\frac{31}{32}$ .

Logically, we can see that the probability of having a tail show up at least once in 5 flips should be pretty high since its probability is $\frac{1}{2}$ , therefore common sense supports our choice.

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