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

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yrozenblum
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In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   29 Oct 2023, 10:45
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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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Re: In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   29 Oct 2023, 12:20
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yrozenblum wrote:
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. \(\frac{1}{3}\)


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 ACurtain BCurtain CCurtain DCurtain ECurtain F
WinningConsolation Consolation WinningConsolation 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
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Re: In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   29 Oct 2023, 13:08
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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

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Re: In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   30 Oct 2023, 22:55
Such types of questions during time crunch is very difficult. Can anyone help to infer that what to do when such Probability + Perm,comb questions come.
I understand the solution and basic concepts but I can not formulate a strategy for Probability questions. The solution is all over the place. Any suggestions?
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Re: In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   08 Apr 2024, 17:53
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yrozenblum wrote:
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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Screenshot 2024-02-24 at 16.12.04.png
­

­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:
WCCWCC --> WCCWCC (1st and 4th are winners both before and after the swap)
CWCCWC --> CWCCWC (2nd and 5th are winners both before and after the swap)
CCWCCW --> CCWCCW (3rd and 6th are winners both before and after the swap)
3 good outcomes.

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

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In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   13 Apr 2024, 13:22
Expert Reply
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.
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Re: In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink] New post   13 Apr 2024, 23:02
Step 1: Total number of arrangements possible: 6!/2!4! = 15
(6 positions, 2 for identical winning prize and 4 for identical consolation prize)

Step 2: Numbers of ways in which the same two curtains have a winning prize behind them

Before arrangement curtain A and curtain D have the winning prize- WCCWCC, after arrangement same two curtains have the winning prize - WCCWCC

Similarly,
A-C, B-W, C-C, D-C, E-W, F-C after arrangement D-C, E-W, F-C, A-C, B-W, C-C
A-C, B-C, C-W, D-C, E-C, F-W after arrangement D-C, E-C, F-W, A-C, B-C, C-W

There are only three ways in which such a scenario would occur. So, the required probability is 3/15 = 1/5
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Re: In a certain TV game show, exactly 1 prize was placed behind each of 6 [#permalink]
13 Apr 2024, 23:02
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