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805+ Level|   Probability|      
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doomedcat
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Ida had 5 cards with matching envelopes - same in design, different in 22 May 2018, 22:44
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

24% (02:13) correct 76% (02:17) wrong based on 265 sessions

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Ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?


A \(\frac{3}{16}\)
B \(\frac{5}{8}\)
C \(\frac{3}{8}\)
D \(\frac{1}{4}\)
E \(\frac{1}{2}\)

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C
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Ida had 5 cards with matching envelopes - same in design, different in 23 May 2018, 00:05
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doomedcat
ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A \(\frac{3}{16}\)
B \(\frac{5}{8}\)
C \(\frac{3}{8}\)
D \(\frac{1}{4}\)
E \(\frac{1}{2}\)

Source : Experts Global
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Hope it helps.
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Ida had 5 cards with matching envelopes - same in design, different in 23 May 2018, 00:01
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doomedcat
ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A \(\frac{3}{16}\)
B \(\frac{5}{8}\)
C \(\frac{3}{8}\)
D \(\frac{1}{4}\)
E \(\frac{1}{2}\)

Source : Experts Global
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Answer: D

The question asks for "exactly one card into the right envelope", hence the combination of right & wrong cards will look like

R W W W W

Assuming the first card is put in right envelope, since there is only one right card out of the 5 for the first envelope, the probability is 1/5

For the second envelope, there are 3 wrong cards out of the 4 cards left, hence the probability for the Second card in wrong envelope is 3/4

For the third envelope, there are 2 wrong cards out of the 3 cards left, hence the probability for the third card in wrong envelope is 2/3

For the fourth envelope, there is 1 wrong card out of the 2 cards left, hence the probability for the fourth card in wrong envelope is 1/2

For the fifth envelope, since only one card is left now & a wrong one at that, hence the probability for the fifth card in wrong envelope is 1

Moreover the arrangement R W W W W can be arranged in 5!/4! ways

Hence the probability of exactly one right card \(= (1/5)*(3/4)*(2/3)*(1/2)*(1) * (5!/4!) = 1/4\)
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Ida had 5 cards with matching envelopes - same in design, different in 15 Apr 2019, 05:54
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Selecting one right envelope = 5 C 1 = 5 ways
Derangment of 4 envelopes = 4!(1- 1/1!+1/2!-1/3!+1/4!) = 9
Total ways of arrangement = 120

Probability of 1 envelope placed in right place = 9*5/120 = 3/8

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Ida had 5 cards with matching envelopes - same in design, different in 15 Apr 2019, 07:32
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Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8[/quote]


This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.[/quote]

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE... :)[/quote]


VeritasKarishma , sayan640
hi could you please advise on the highlighted part. How did we arrive at 9?
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Ida had 5 cards with matching envelopes - same in design, different in 16 Aug 2020, 07:20
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Ida had 5 cards with matching envelopes - same in design, different in color.

=> Cards : \(C_1\), \(C_2\),\(C_3\),\(C_4\),\(C_5\).

=> Envelopes: \(E_1\), \(E_2\),\(E_3\),\(E_4\),\(E_5\).(Same design, different color).

The probability that exactly one card got into the matching envelope

=> Probability : \(\frac{Desired results }{ Total number of results.}\)

5 cards can be put into 5! ways = 120.

Exact one card correct:

Example : \(C_1\) goes in \(E_1\) (Correct).

=> No of ways one card can be chosen which go into the correct envelope out of 5 cards is = \(^{5}\mathrm{C_1}\) = 5
= 5.

No of ways \(C_2\) goes into the wrong envelope: \(E_3\), \(E_4\), and \(E_5\) = 3 ways - (Suppose \(C_2\) has been put in \(E_3\)).

No of ways \(C_3\) goes into the wrong envelope: \(E_2\), \(E_4\), and \(E_5\) = 3 ways - (Suppose \(C_3\) has been put in \(E_2\)).

No of ways \(C_4\) goes into the wrong envelope: \(E_5\) = 1 way

No of ways \(C_5\) goes into the wrong envelope: \(E_4\) = 1 way

Total ways : 3 * 3 * 1 * 1 = 9

Desired result: 5 * 9

Total results: 120

Probability: \(\frac{5 * 9 }{ 120}\) = \(\frac{45 }{ 120}\) = \(\frac{3}{8}\)

Answer C
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Ida had 5 cards with matching envelopes - same in design, different in 17 Aug 2023, 08:30
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The number of ways 1 card finds the correct envelope is:

Total ways-3 correct -2 correct - 0 correct

Total ways is 5!=120

3 correct: 5!/3!2! = 10 ways the 3 correctly inserted can be selected × 1 way the remaining 2 can be inserted incorrectly= 10

2 correct: 5!/2!3! = 10 ways the 2 correct can be selected. The remaining 3 can be seen quickly to have only 2 ways to be inserted incorrectly, eg, A can only be inserted into B or C = 20

0 ways, the difficult one. We know A has 4 incorrect choices and that the total calculation is a multiplication so that the answer is an integer multiple of 4, call it

4X

So the answer must be

(120-1-10-20-4X)/120 = (89-4X)/120

Testing the first answer for example, can 89-4X= 3/16 *120 ? No, because 120 isn't a multiple of 16.

Can it equal 5/8*120 ?

5/8 * 120 = 75

89-75=14 is not a multiple of 4

How about 3/8 ? 3/8 * 120 =45
89-45=44 is a multiple of 4
3/8 is the correct answer

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Ida had 5 cards with matching envelopes - same in design, different in 28 Sep 2025, 20:13
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We need to carefully recheck.

Total arrangements = 5! = 120.

Let’s calculate favorable cases (exactly 1 card in the correct envelope).

Step 1: Choose the 1 card that is correctly placed.
5 ways.

Step 2: Arrange the remaining 4 cards so that none of them goes into its correct envelope (derangement of 4).
Number of derangements of 4 = !4 = 9.

So favorable = 5 × 9 = 45.

Step 3: Probability
= 45 / 120
= 3 / 8.

Answer: C.
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