Terry holds 12 cards, each of which is red, white, green, or blue. If…

Reading Time: 2 minutes

Last Updated on May 6, 2023

GMAT OFFICIAL GUIDE DS

Solution:

We are given that Terry has 12 total cards. The cards are colored red, white, green, or blue. We need to determine whether the probability of selecting a red or a white card is less than ½. Remember, since we are determining the probability of selecting a red or a white card, we must add the probabilities.

Is P(red card) + P(white card) < ½?

Since the sum of all probabilities in a sample set is equal to 1, we also know that:

P(red card) + P(white card) + P(blue card) + P(green card) = 1

P(red card) + P(white card) = 1 – [P(blue card) + P(green card)]

Thus, if we can determine the sum of the probabilities of selecting a red card and of selecting a white card OR the sum of the probabilities of selecting a blue card and of selecting a green card, we also could determine the probability of selecting a red or white card.

Statement One Alone:

The probability that the person will select a blue card is 1/3.

Since we don’t know the probability of selecting a green card, we cannot determine:

1 – [P(blue card) + P(green card)] OR

P(red card) + P(white card)

Thus, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The probability that the person will select a red card is 1/6.

Since we don’t know the probability of selecting a white card, we cannot determine:

P(red card) + P(white card)

Thus, statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using both statements together we know the following:

P(red card) = 1/6

P(blue card) = 1/3

Substituting this into our two expressions we have:

P(red card) + P(white card) = 1/6 + P(white card) = ?

1 – [1/3 + P(green card)] = ?

We see that we still do not have enough information to determine whether the probability of selecting a red card or a white card is less than ½.

Answer: E

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