Not all probability questions on the GMAT will be about dice, playing cards and marbles. Here are examples of other possible scenarios.

Not all probability questions on the GMAT will be about dice, playing cards and marbles. Here are examples of other possible scenarios.

A study found that 80% of drivers, who ride with a passenger, wear seatbelts. Passengers wear seatbelts 70% of the time when the driver wears a seatbelt, and 55% of the time when the driver doesn’t wear a seatbelt. What is the probability that a passenger wears a seatbelt?

Solution

You can use a tree diagram to visualize the possibilities. Write the probabilities on each branch.


There are 2 paths that give the desired outcome of a passenger wearing a seatbelt. Find and add the probabilities of each path.

(Path 1) Driver wears and passenger wears: 80% × 70% = 56%
(Path 2) Driver does not wear and passenger wears: 20% × 55% = 11%

The probability is 56% + 11% = 67%.

A passenger wears a seatbelt 67% or about \dfrac{2}{3} of the time.

In a 1-mile race, three different schools (Washington High, Duke High and Cherry Hill High) each have 5 competitors. What is the probability that a student from Cherry Hill will take first place, a student from Duke will take second place and another student from Duke will take third place?

Solution

This question has both “or” and “and” in the same problem: independent and dependent.

Since 2 students from Duke are desired outcomes, those are dependent events.

Find each probability.

Cherry Hill student first: \dfrac{5}{15} = \dfrac{1}{3}
Duke student takes second: \dfrac{5}{14}
Duke student takes third: \dfrac{4}{13}

Since this is an “and” question, multiply the probabilities. Don’t forget to factor and reduce before you multiply.

\dfrac{1}{3} × \dfrac{5}{14} × \dfrac{4}{13} = \dfrac{5 × 2 × 2}{3 × 2 × 7 × 13} = \dfrac{10}{273}

At 3 p.m. Jennifer went into labor. There is a 0.7 chance her baby will be born during each hour that she is in labor. What is the probability that her baby was born at 5:30 p.m. on the same day?

A) 0.027
B) 0.063
C) 0.147
D) 0.27
E) 0.343

Solution

This question is an independent probability question in disguise. The probability that her baby will be born each hour does not change. Each hour, there is a 0.7 chance the baby will be born, which means there is a 0.3 chance the baby will not be born.

Method 1

From 3 p.m. to 4 p.m., the baby was not born and the probability of that happening is 0.3.
From 4 p.m. to 5 p.m. the probability also is 0.3.
From 5 p.m. to 6 p.m., when the baby was born, the probability is 0.7.
Since the baby must not be born in the first or second hour, and must be born in the third hour, this is an “and” question, so the probability is (0.3)(0.3)(0.7) = 0.063

The correct answer is choice (B).

Method 2

You can use a tree diagram to visualize the possibilities. Write the probabilities on each branch.


There is just one path that gives the desired outcome.

(0.3)(0.3)(0.7) = 0.063

The correct answer is choice (B).

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