In a certain TV game show, exactly 1 prize was placed behind each of 6

Question

In a certain TV game show, exactly 1 prize was placed behind each of 6 curtains, which were labeled A, B, C, D, E, and F. The correspondance among curtains and prizes was random. Of the prizes, 4 were consolation prizes and 2 were winning prizes. Later, the prizes were rearranged, so that the prizes that had been behind Curtains A, B, and C were now behind Curtains D, E, and F, respectively, and the prizes that had been behind Curtains D, E, and F were now behind Curtains A, B, and C, respectively. What is the probability that the same 2 curtains had a winning prize behind them before and after the rearrangement of the prizes?

A. \(\frac{1}{12}\)

B. \(\frac{1}{6}\)

C. \(\frac{1}{5}\)

D. \(\frac{1}{4}\)

E. \(\frac{1}{3}\)

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Official Answer

C

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gmatophobia · Accepted Answer

Let's assume the prizes are indicated as shown below -

Winning Prize =  W_1  &  W_2

Consolidation Prize =  C_1, C_2, C_3,  &  C_4

Note: The prizes are identical.

Only when two curtains that are being swapped have winning prizes before the swap, will the curtains have winning prizes after the swap. For example, if there is a winning prize behind curtain A, and curtain D before the swap, there will be a winning prize behind the two curtains after the swap. In all other possibilities, the type of prize will not be retained.

Let's assume curtain A and curtain D have the winning prize.

The number of ways of arranging two winning prizes and four consolidation prizes in a single row =  \frac{6!}{2!*4!} = 15

Of the 15 arrangements only one arrangement indicates that curtain A and curtain D have the winning prize (shown below)-
 
   Curtain A    Curtain B    Curtain C    Curtain D    Curtain E    Curtain F   
  Winning  Consolation   Consolation   Winning  Consolation   Consolation   
 
Therefore, the probability of (curtain A and curtain D having the winning prize) =  \frac{1}{15}

The same analysis holds true to calculate the probability of curtain B and curtain E having the winning prize and of curtain C and curtain F having the winning prize.

The probability of (curtain B and curtain E having the winning prize) =  \frac{1}{15}

The probability of (curtain C and curtain F having the winning prize) =  \frac{1}{15}

Total Probability = The probability of (curtain A and curtain D having the winning prize)  OR  The probability of (curtain B and curtain E having the winning prize)  OR  The probability of (curtain C and curtain F having the winning prize)

Total Probability =  \frac{1}{15} * 3 = \frac{1}{5}

Option C

Regor60 · Answer

Both prizes need to be behind a pair being switched both before and after the switch.

Further, the prizes are indistinguishable from one another.

Put one prize behind one door.

The probability that the 2nd prize is placed behind the paired door is therefore

1/5

Another way:

Number of ways to select 2 doors behind which to place the prizes:

6!/2!4! = 15

The number of door pairs that match the swapping condition:

3

Probability of a swapping door pair being selected from all possible door pairs:

3/15 = 1/5

Posted from my mobile device

GMATGuruNY · Answer

P =  \frac{good-outcomes}{all-possible-outcomes}

All possible outcomes:
Among the 6 curtains, two must be winners.
From 6 curtains, the number of ways to choose 2 to be winners = 6C2  = \frac{6*5}{2*1} = 15

Let W = winning curtain and C = consolation curtain

Good outcomes:
When the first 3 curtains swap contents with the last 3, the same pair must be winners both before and after the swap.
Only 3 cases satisfy this condition:
 WCC  WCC  -->  WCC  WCC  (1st and 4th are winners both before and after the swap)
 CWC  CWC - ->  CWC  CWC  (2nd and 5th are winners both before and after the swap)
 CCW  CCW  -->  CCW  CCW  (3rd and 6th are winners both before and after the swap)
3 good outcomes.

P =  \frac{good}{all} = \frac{3}{15} = \frac{1}{5}

C

Vinit800HBS · Answer

A different and simpler approach can be this.

Select any 1 of the 6 curtains.

Now, every curtain has a unique pair with another curtain (A with D, B with E, C with F).

So, of the remaining 5 curtains, only 1 curtain will pair up with the first curtain that we have selected.

Hence, probability will be 1/5 to get both the winning prizes.

korteraz · Answer

My solution:
The probability that the winning prize is behind one of the curtains is 2/6 (let's say A).
The probability that another winning prize is behind another specific curtain is 1/5 (let's say D).

The probability of these two together is 2/6 * 1/5= 2/30.

There are 3 scenarios for this to happen so 3*2/30 = 1/5 (AD, BE, CF)

GMATinsight · Answer

it’s possible only when either
A and D had Winning prizes
OR
B and E had Winning prizes
OR
C and F had Winning prizes

i.e. Favorable outcomes = 3

Total Outcomes = 6C2 = 15 (choose two curtain behind which winning prizes may be kept)

Required Probability = 3/15 = 1/5

Answer: Option C

pranjalshah · Answer

Prizes don't necessarily have to be identical. Let's say we have W1W2C1C2C3C4.

Total possible ways to arrange them = 6!

For the winning prize to be behind the same curtain before and after the swap, they must be at their respective mirrored spots, for example:

W1 _ _ | W2 _ _ or _ W1 _ | _ W2 _ or _ _ W1 | _ _ W2

Total no of ways we can do this is 3*2. As you can see above, we have 3 ways to put w1 and w2. Also, we can interchange w1 and w2 for ex. 
W2 _ _ | W1 _ _

Therefore we have 6 ways to put the winning prizes. For the remaining 4 positions, we can put C1C2C3C4 in any way, which further gives us 4! combinations.

Therefore total favourable ways = 6*4!

Probability = favorable ways/total ways

= (6*4!)/6!

= 1/5

Note: The answer would be same if you consider them identical. Total favorable ways = 3 (W_ _ | W_ _ or _W_ | _W_ or _ _W | _ _W)
Total possible ways = 6!/(2!*4!) = 5*3

Probability = 3/(5*3) = 1/5

KarishmaB · Answer

Responding to a pm: Visualise and divide the arena into ABC | DEF A's prize is being switched with D's prize, B's with E's and C's with F's. Think about when is it possible that the same 2 curtains had a winning prize behind them before and after the rearrangement of the prizes? If both winning prizes are in ABC, will flipping keep the prizes same? No. If both are in DEF, then again it is not possible. So 1 prize must be in ABC and the other in DEF. For same two curtains to have the winning prize after the flip too, the winning prizes must be in the paired curtains. So either AD or BE or CF should have both winning prizes. Hence there are only 3 ways of putting the winning prizes. All other curtains have consolation prizes. In how many total ways can we arrange the 6 prizes? In 6!/2!*4! = 15 ways (since 4 consolation prizes are same and 2 winning prizes are same) Hence the probability = 3/15 = 1/5 Answer (C) Probability discussed here: https://youtu.be/0BCqnD2r-kY

Pratyaksh1907 · Answer

Hi Experts!
I didn't find any approach like mine for this Q. Please check if this is the right way to do it or not.
Thanks

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Need help to validate my process. Haven't found any approach like mine for this Q. Please review it.
Thanks

In a certain TV game show, exactly 1 prize was placed behind each of 6

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Curtain A	Curtain B	Curtain C	Curtain D	Curtain E	Curtain F
Winning	Consolation	Consolation	Winning	Consolation	Consolation

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In a certain TV game show, exactly 1 prize was placed behind each of 6

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