At a blind taste competition, a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of

[M7MBA]

At a blind taste competition, a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. \(\dfrac1{12}\)

B. \(\dfrac5{14}\)

C. \(\dfrac49\)

D. \(\dfrac12\)

E. \(\dfrac23\)

Answer: B

Source: GMAT Club Tests

[Brent@GMATPrepNow]

At a blind taste competition, a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. \(\dfrac1{12}\)

B. \(\dfrac5{14}\)

C. \(\dfrac49\)

D. \(\dfrac12\)

E. \(\dfrac23\)

Answer: B

Source: GMAT Club Tests

This question can be solved using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(contestant does not taste all 3 samples) = 1 - P(contestant DOES taste all 3 samples)

P(contestant DOES taste all 3 samples)
For this event to occur, the contestant must taste 2 cups of one sample, 1 cup from another sample, and 1 cup from another sample.

Let's take the task of tasting all 3 samples and break it into STAGES.
Stage 1: Select the tea that will be tasted twice. There are 3 types of tea, so stage 1 can be completed in 3 ways.

Stage 2: Choose 2 cups to taste from tea selected in stage 1. Since the order in which we select the 2 cups does not matter, we can use combinations. We can select 2 cups from 3 cups in 3C2 ways(= 3 ways).

Stage 3: From one of the two remaining (untasted) teas, select 1 cup to taste. There are 3 cups, so stage 3 can be completed in 3 ways.

Stage 4: Select 1 cup from the last remaining (untasted) tea. There are 3 cups, so stage 4 can be completed in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages in (3)(3)(3)(3) ways (= 81 ways)

The TOTAL number of ways to select 4 cups from 9 cups = 9C4 = 126

So, P(contestant DOES taste all 3 samples) = 81/126 = 9/14

This means that P(contestant does not taste all 3 samples) = 1 - 9/14
= 5/14
= B

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Cheers,
Brent

[Brent@GMATPrepNow]

At a blind taste competition, a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. \(\dfrac1{12}\)

B. \(\dfrac5{14}\)

C. \(\dfrac49\)

D. \(\dfrac12\)

E. \(\dfrac23\)

Answer: B

Source: GMAT Club Tests

Let's say that there are 3 kinds of tea: A, B and C, and there are 3 cups of each tea.

First find P(contestant does not taste tea A)
P(contestant does not taste tea A) = P(1st selection is not tea A AND 2st selection is not tea A AND 3rd selection is not tea A AND 4th selection is not tea A)
= P(1st selection is not tea A) x P(2st selection is not tea A) x P(3rd selection is not tea A) x P(4th selection is not tea A)
= 6/9 x 5/8 x 4/7 x 3/6
= 5/42

Now find P(contestant does not taste tea B)
The steps to find this probability will be the same as steps taken to find the above probability.
So, P(contestant does not taste tea B) = 5/42

Likewise, P(contestant does not taste tea C) = 5/42

So, P(contestant does not taste all of the samples) = 5/42 + 5/42 + 5/42
= 5/14
= B
Cheers,
Brent