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3 votes
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A box contains 10 mangoes out of which 4 are rotten  .Two mangoes are taken out together. If one of them is found to be good ,then find the probability other is also good

3 Answers

7 votes
7 votes

Probability that both mango is rotton = 4c2/10c2 [ i.e. 4 are rotton u choose any two of them in 4c2 ways]

Probability that atleast of of them is good = Total [since total probability is 1]  - Probability that both mango is rotton

                                                            = 1 - 4c2/10c2

                                                             = 13/15

Probability that both mango is good = 10 - 4 = 6 choose any 2 mango out of these 6 = 6c2/ 10c2 =1/3

So required probability [ Both are good ] =   15/13 * 1/3

                                                           = 5/13 [BAYES THEOREM p(B/A) = p(A intersection B) /P(A) ]

3 votes
3 votes

Here is the solution

1 votes
1 votes
At least one mango is good

=1-4c2/10c2

= 13/15 say A

Both are good =6c2/10c2 = 1/3 say B

So required probability is B/A

(1/3) *(15/13) = 5/13

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