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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
26/43  ?

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) ]

answered by Veteran (52.8k points) 21 71 325

### P(A)=P(Ist mango being good)=6C1/10C1 right?

Here is the solution

answered by Veteran (30.2k points) 107 371 691
why u consider at least one is good,in question it is given one of them is good ,so m' should be p(one of them is good).if a am wrong please correct me
+1 vote
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
answered by Veteran (25.3k points) 13 60 193
in question one of them is good,then why you consider at least one is good