Find the probability that the letter to A is in the correct envelope, the letter to B is in an incorrect envelope.

Transcription

1 A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addresses the envelopes, at random, one each to A, B, C, D and E. (i) (ii) (iii) Find the probability that the letter to A is in the correct envelope and the letter to B is in an incorrect envelope. Find the probability that the letter to A is in the correct envelope, given that the letter to B is in an incorrect envelope. Find the probability that both letters to A and B are in incorrect envelopes. Solution: (i) Probability Method: The scenario can be viewed as distributing the letters to the Envelope one by one (without replacement of course), starting with Envelope A. Required Probability = = 3 st Letter goes to Envelope A (5 letters to choose from in total, only correct letter) (4 letters to choose from in total after distributing to A, incorrect letters to B have 3 choices: C, D and E) Number of desired outcomes (Envelope A contains the correct letter and Envelope B the incorrect letter) = ( 3 ) 3! st Letter goes to Envelope A (only way to do that) 3 ways for an incorrect letter to go into Envelope B (choosing from Letter C, D and E) The number of ways to arrange the remaining 3 letters in any order without restriction Required Probability = (3 ) 3! 5! = 3 Total number of ways to arrange 5 letters amongst the 5 envelopes in any order

2 (ii) Probability Method: This is a Conditional Probability Question. Probability of Given event = P(letter to B is in an incorrect envelope) = incorrect letters for Envelope B (Letter A, C, D and E) Probability of Intersection Event = P(letter to B is in an incorrect envelope AND A in the correct Envelope) = 3 (Found in part (i)) Required Probability = Probability of Intersection Event Probability of Given event = ( 3 ) ( 4 5 ) = 3 6 Required Probability = Number of Ways for A in Correct Letter AND Envelope B with incorrect Letter Envelope B with incorrect letter Number of ways Envelope B contains incorrect letter = ( 4 ) 4! = 96 Number of Ways for A in Correct Letter AND Envelope B with incorrect Letter = ( 3 ) 3! = 8 Required Probability = 8 96 = 3 6

3 (iii) This part is the challenging one. Direct Approach: The main challenge is the ability to identify the need to break this question into cases. For letters A and B to be placed in the incorrect envelopes. There are 4 mutually exclusive (nonoverlapping) cases. Case I: Letters A and B are swap with each other. It does not matter how the rest of the 3 envelopes gets their letters. Case II: Letter A goes into Envelope C, D and E and Letter B goes into Envelope A. Case III: Letter B goes into Envelope C, D and E and Letter A goes into Envelope B. Case IV: Both letters A and B goes into Envelope C, D and E. Probability Method: Case : 5 4 = Case 2: = 3 Case 3: Same as Case 2 = 3 Case 4: = 6 Letter A goes into Envelope C, D and E (out from the available 5 possible envelope of choices) Letter B goes into 2 of the remaining Envelope C, D and E (out from the available 4 possible envelope of choices AFTER Letter A gets distributed into the Envelope ) Required Probability = = 3 Case = 3! = 6 Case 2: ( 3 ) 3! = 8

4 way for Letter A to go into Envelope B Case 3: ( 3 ) 3! = 8 No. of ways Letter B can be distributed into Envelope C, D and E No. of ways the remaining 3 letters can be distributed amongst the 3 Envelopes Case 4: ( 3 ) 2! 3! = 36 2 Total Number of ways 5 letters get distributed to the 5 Envelopes = 5! Required Probability = ! = 78 = 3 Indirect Approach: The indirect approach here is to rephrase the problem into Set and approach the question using Set Notation formula. A cue for such a method is the involvement of the phrase OR or the multiple conditions involving 2 or more NOT events (like in this instance, the 2 conditions are 2 letters NOT in the correct Envelopes). We shall them define the 2 events as follows: Let: A = Envelope A contains the correct Envelope B = Envelope B contains the correct Envelope Then the desired outcome will be = A B Since P(A B ) can be calculated easily using De Morgan s formula, we shall apply that in the computation of P(A B ) P(A B ) = P[(A B) ] = P(A B) P(A B) = P(A) + P(B) P(A B) P(A) = 5, P(B) = 5 [Probability that any letter can distributed correctly into their Envelopes with no other restrictions imposed is the same]

5 P(A B) = 5 4 = Required Probability = [P(A) + P(B) P(A B)] = [ ] = 3