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Carcass has 4 different letters to be sent to 4 different addresses.
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15 Sep 2021, 21:45
15 Sep 2021, 21:45
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Question Stats:
91% (01:51) correct
8% (01:18) wrong based on 12 sessions
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Carcass has 4 different letters to be sent to 4 different addresses. For each letter, he prepared an envelope with its correct address. He then put 4 letters into the 4 prepared envelopes at random.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Probability that only 1 letter was put into the envelope with its correct address |
\(\frac{2}{5}\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: Carcass has 4 different letters to be sent to 4 different addresses.
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16 Sep 2021, 06:54
16 Sep 2021, 06:54
1
Here is an inefficient way of solving this:
Let 4 envelopes be E1,E2, E3 and E4 and 4 letters be L1,L2,L3 and L4.
The total number of ways to be matched is 4! = 24.
And there could be, only one matching, two matching, three matching and all four matching.
Only one matching for just E1 and L1 will be as follows:
E1 E2 E3 E4
L1 L3 L4 L2
L1 L4 L2 L3
Just 2 ways for E1 L1, similarly for other 3 including will be 4 * 2. so P = 8/24= 1/3
Answer is B) 2/5 is greater.
Let 4 envelopes be E1,E2, E3 and E4 and 4 letters be L1,L2,L3 and L4.
The total number of ways to be matched is 4! = 24.
And there could be, only one matching, two matching, three matching and all four matching.
Only one matching for just E1 and L1 will be as follows:
E1 E2 E3 E4
L1 L3 L4 L2
L1 L4 L2 L3
Just 2 ways for E1 L1, similarly for other 3 including will be 4 * 2. so P = 8/24= 1/3
Answer is B) 2/5 is greater.
Re: Carcass has 4 different letters to be sent to 4 different addresses.
[#permalink]
16 Sep 2021, 21:34
16 Sep 2021, 21:34
1
Put correct in 1st envelope, 3 options to put in 2nd envelope only 2 of which are incorrect and then only 1 incorrect for third. Total cases 2*1=2 when 1st envelope is correct
Repeat for 4envelopes= 2+2+2+2=8
Total case = 4!=24
Probability = 8/24=1/3<2/5
Repeat for 4envelopes= 2+2+2+2=8
Total case = 4!=24
Probability = 8/24=1/3<2/5
