GMAT Quantitative - Calculating probability

Sheryl is competing in an archery tournament. She gets to shoot three arrows at a target, and her best one counts.

Sheryl hits the bullseye 42% of the time. What is the probability (two decimal places) that she will hit the bullseye at least once in her three tries?

Explanation

This is most easily solved by finding the probability that she will not hit the bullseye at all in her three tries. If she hits 42% of the time, she misses 58% of the time, and the probability she misses three times will be

$0.58 ^{3} = 0.195112 \approx 0.20$ .

The probability of hitting the bullseye at least once in three tries is the complement of this, or .

Two dice are rolled. What is the probability that the sum of both dice is greater than 8?

$\small \frac{5}{18}$

$\small \frac{7}{18}$

$\small \frac{1}{8}$

$\small \frac{5}{9}$

Explanation

There are 36 possible outcomes ( $6\times 6=36$ ). 10 out of the 36 outcomes are greater than 8: (6 and 3)(6 and 4)(6 and 5)(6 and 6)(5 and 4)(5 and 5)(5 and 6)(4 and 5)(4 and 6)(3 and 6).

$\small \frac{10}{36}\ =\ \frac{5}{18}$

It costs $10 to buy a ticket to a charity raffle in which three prizes are given - the grand prize is $3,000, the second prize is $1,000, and the third prize is $500. Assuming that all of 1,000 tickets are sold, what is the expected value of one ticket to someone who purchases it?

Explanation

If 1,000 tickets are sold at $10 apiece, then $10,000 will be raised. The prizes are $3,000, $1,000, and $500, so $4,500 will be given back, meaning that the 1,000 ticket purchasers will collectively lose $5,500. This means that on the average, one ticket will be worth

$\frac{-$5,500}{1,000} = -$5.50$

This is the expected value of one ticket.

What is the probability of rolling an even number on a standard dice?

Explanation

A standard dice has 6 faces numbered .

There are even numbers, , divided by the total number of faces:

$\frac{3}{6}= 0.5$

How many even four-digit numbers larger than 4999 can be formed from the numbers 2, 4, 5, and 7 if each number can be used more than once?

Explanation

Since the number must be larger than 4999, the thousand’s digit has to be 5 or 7. We are also told that the number must be even. Thus, the unit’s digit must be 2 or 4. The middle digits can by any of the numbers 2,4,5, or 7. Therefore, we have a total of $2\cdot 4\cdot 4\cdot 2=64$ possibilities.

What is the probability of sequentially drawing 3 aces from a deck or regular playing cards when the selected cards are not replaced?

$\frac{1}{5525}$

$\frac{1}{2197}$

$\frac{1}{64}$

$\frac{1}{140,608}$

Explanation

The probability of drawing an ace first is $\frac{4}{52}$ or $\frac{1}{13}$ .

Assuming an ace is the first card selected, the probability of selecting another ace is $\frac{3}{51}$ or $\frac{1}{17}$ .

For the third card, the probability is $\frac{2}{50}$ or $\frac{1}{25}$ .

To calculate the probability of all 3 events happening, you must multiply the probabilities:

$\frac{1}{13}\times\frac{1}{17}\times\frac{1}{25}=\frac{1}{5525}$

A drawer has 4 green socks, 6 blue socks, 12 white socks, 8 black socks, and 2 pink socks. If you reach in and pull out a sock at random, what is the probability the sock will be blue?

$\frac{3}{16}$

$\frac{3}{8}$

$\frac{1}{6}$

$\frac{3}{32}$

$\frac{5}{16}$

Explanation

If you reach in and pull out a sock at random, the probability of it being blue is equal to the number of blue socks divided by the total number of socks in the drawer. First we'll calculate the total number of socks in the drawer:

Now we divide the number of blue socks by the total number of socks to find the probability of picking a blue sock:

$P=\frac{6}{32}=\frac{3}{16}$

Among a group of 300 people, 15% play soccer, 21% play baseball, and 9% play both soccer and baseball. If one person is randomly selected, what is the probability that the person selected will be one who plays baseball but NOT soccer?

$\frac{3}{25}$

$\frac{9}{50}$

$\frac{4}{7}$

$\frac{21}{100}$

Explanation

Since there are 300 people, $300\cdot .21=63$ people play baseball and $300\cdot .09=27$ of those people play both baseball and soccer. Therefore, there are people who play baseball but not soccer.

Probability: $\frac{36}{300}=\frac{3}{25}$

If you flip a quarter three times, what is the probability of it landing on heads all three times?

$\frac{1}{8}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{16}$

Explanation

Every time you flip a coin, there is a 1 in 2 chance of it landing on heads. So, if we want to know the probability of a coin landing on heads a certain number of times times in a row, we multiply the probability of that occurrence for however many times the coin is flipped. For three flips, this gives us:

$P=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$

So there is a 1 in 8 chance that a coin will land on heads three times in a row.

A jar contains 8 blue marbles and 4 red marbles. What is the probability of picking a blue marble followed by a red marble if the first marble chosen is not put back in the jar?

$\frac{8}{33}$

$\frac{2}{9}$

$\frac{2}{11}$

$\frac{23}{66}$

$\frac{3}{4}$

Explanation

There are 12 marbles total. The probability of picking a blue marble first is $\frac{8}{12}=\frac{2}{3}$ . The probability of then picking a red marble out of the 11 remaining marbles is $\frac{4}{11}$ . Therefore, the probability is $\frac{2}{3} \times\frac{4}{11}=\frac{8}{33}$ .

Calculating probability

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