GMAT Math : Discrete Probability

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Example Question #295 : Arithmetic

We toss a coin times, what is the probabilty of obtaining a head on all tosses?

Possible Answers:

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{1}{32}$

$\frac{1}{2}$

$\frac{1}{8}$

Correct answer:

$\frac{1}{8}$

Explanation:

This simple problem asks us to find the probability of an event occuring 3 times.

A head shows up with a probability of $\frac{1}{2}$ , since we throw the coin thrice, the final answer is given by $\frac{1}{2^{3}}$ or $\frac{1}{8}$ .

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Example Question #51 : Calculating Discrete Probability

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

Possible Answers:

$\frac{1}{32}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{2}$

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.

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Example Question #291 : Arithmetic

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

Possible Answers:

$\frac{27}{32}$

$\frac{3}{4}$

$\frac{29}{32}$

$\frac{1}{2}$

$\frac{31}{32}$

Correct answer:

$\frac{31}{32}$

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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Example Question #41 : Probability

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

Possible Answers:

$\frac{9}{26}$

$\frac{3}{52}$

$\frac{156}{2704}$

$\frac{11}{26}$

$\frac{25}{52}$

Correct answer:

$\frac{11}{26}$

Explanation:

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

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Example Question #41 : How To Find The Probability Of An Outcome

A coin is flipped four times. What is the probability of getting heads at least three times?

Possible Answers:

$\frac{1}{16}$

$\frac{3}{4}$

$\frac{11}{16}$

$\frac{5}{16}$

$\frac{1}{4}$

Correct answer:

$\frac{5}{16}$

Explanation:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

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Example Question #51 : Calculating Discrete Probability

Rolling a four-sided dice, what is the probability of rolling a three times out of four?

Possible Answers:

$\frac{3}{128}$

$\frac{3}{64}$

$\frac{243}{256}$

$\frac{13}{256}$

$\frac{3}{256}$

Correct answer:

$\frac{3}{64}$

Explanation:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case rolling a four), and is the probability of success (one in four).

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{4})^3(1-\frac{1}{4})^{4-3}=4(\frac{1}{64})(\frac{3}{4})=\frac{3}{64}$

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Example Question #47 : Probability

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

Possible Answers:

$\frac{6}{7}$

$\frac{1}{128}$

$\frac{255}{256}$

$\frac{1}{7}$

$\frac{127}{128}$

Correct answer:

$\frac{127}{128}$

Explanation:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

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Example Question #41 : How To Find The Probability Of An Outcome

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

Possible Answers:

$\frac{2}{14}$

$\frac{5}{7}$

$\frac{4}{7}$

$\frac{3}{14}$

$\frac{5}{14}$

Correct answer:

$\frac{5}{14}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

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Example Question #41 : How To Find The Probability Of An Outcome

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?

Possible Answers:

$\frac{6}{7}$

$\frac{1}{7}$

$\frac{5}{13}$

$\frac{5}{14}$

$\frac{17}{14}$

Correct answer:

$\frac{6}{7}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)$

The interesection is:

$P(A\bigcap B)=P(A)P(B)=\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

So, we can find the probability of drawing at least one consonant:

$P(A\bigcup B)= \frac{5}{7}+\frac{1}{2}-\frac{5}{14}=\frac{10}{14}+\frac{7}{14}-\frac{5}{14}=\frac{6}{7}$

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Example Question #2871 : Act Math

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is drawn from Set A, and one from Set B. What is the probability of drawing a matching pair of letters?

Possible Answers:

$\frac{1}{21}$

$\frac{1}{42}$

$\frac{4}{13}$

$\frac{1}{7}$

$\frac{2}{13}$

Correct answer:

$\frac{1}{21}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B, 7(6)=42 .

Therefore, the probability of drawing a matching set is:

$\frac{2}{42}=\frac{1}{21}$

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GMAT Math : Discrete Probability

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #295 : Arithmetic

Example Question #51 : Calculating Discrete Probability

Example Question #291 : Arithmetic

Example Question #41 : Probability

Example Question #41 : How To Find The Probability Of An Outcome

Example Question #51 : Calculating Discrete Probability

Example Question #47 : Probability

Example Question #41 : How To Find The Probability Of An Outcome

Example Question #41 : How To Find The Probability Of An Outcome

Example Question #2871 : Act Math

All GMAT Math Resources

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