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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

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 

Grade:12

4 Answers

Puneet Mittal AskiitiansExpert-IITD
22 Points
13 years ago

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

swetha viriyala
31 Points
13 years ago

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.. 

Ankur Rana
35 Points
7 years ago
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) ]
Ankur Rana
35 Points
7 years ago
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) ]

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