One box contains 10 mangoes. 4 of them are rotten and remaining all are good mangoes. 2 mangoes are selected at random. one of them is good one. Find the probability that the second one is also a good one.

Please answer it fastly

HI Swetha,

The box  has 4 rotten mangoes and 6 good mangoes.

Now probability of selecting one good mango from the box is number of good mangoes by total number of mangoes.

that is P1 =  6 / 10. 

After selecting one good mango, we have 9 mangoes left. 4 are rotten and 5 are good mangoes.

Now probability of selecting one good mango is 5/9.

Thus P2 = 5/9

Total probability P = P1*P2 =  (6/10) * (5/9)

Please feel free to ask as many questions you have.

Puneet

Sir....thank you so much for answering the question but given that the answer is 5/13 is it true or false ? can you please say.. 

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 = 13/15Probability that both mango is good = 10 - 4 = 6 choose any 2 mango out of these 6 = 6c2/ 10c2 =1/3So required probability [ Both are good ] = 15/13 * 1/3 = 5/13 [BAYES THEOREM p(B/A) = p(A intersection B) /P(A) ]

Probabiliy 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 = 13/15Probability that both mango is good = 10 - 4 = 6 choose any 2 mango out of these 6 = 6c2/ 10c2 =1/3So required probability [ Both are good ] = 15/13 * 1/3 = 5/13 [BAYES THEOREM p(B/A) = p(A intersection B) /P(A) ]