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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
asked in Probability by Active (1.5k points)   | 237 views
26/43  ?

3 Answers

+5 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) ]

 

 

 

answered by Veteran (44.8k points)  

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

 

+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 (22.2k points)  
in question one of them is good,then why you consider at least one is good
+1 vote

Here is the solution

answered by Veteran (24.2k points)  
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


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