MadeEasy Workbook: Probability - Conditional Probability

Question

Q : Two computers A and B are to be sold . A salesman who is assigned the job of selling these has the chances of 60 percent and 40 percent respectively to get success.The two computers may be sold independently.Given that at least one computer has been sold , the probability that computer A has been sold is ..........(Round off your answer correct to 2 decimal places).

Comments

Can you plz elaborate how it is 3/5??

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is ans 0.789?

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yes chk

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Let X be the event that "atleast one has been sold"

And Y be the event that "A has been sold"

the question asks conditional probability pr{Y/X}

so pr{Y/X} = pr{X $\bigcap$ Y} / pr{X}

pr{X} = 0.6*0.6+ 0.4*0.4 + 0.6*0.4 = 0.76

pr{X $\bigcap$ Y} = 0.6*0.6 + 0.6*0.4 = 0.60

so pr{Y/X} = pr{X $\bigcap$ Y} / pr{X} = 0.60/0.76

Answer 1

Here it is given atleast 1 computer is sold.

Now, we have to find the probability that A is sold

So, here 3 chances

1) A sold

2) B sold

3) both sold

So, among these 3 chances probability of A sold= $\frac{1}{3}\times \frac{40}{100}$

Probability of B sold=$\frac{1}{3}\times \frac{60}{100}$

Probability of both sold=$\frac{1}{3}\times 1$

So,total probability= $\frac{\frac{1}{3}\times \frac{40}{100}+\frac{1}{3}\times 1}{\frac{1}{3}\times \frac{40}{100}+\frac{1}{3}\times \frac{60}{100}+\frac{1}{3}\times 1}$=$\frac{7}{10}$

Comments

@ yashgupta1992

See the solution of @Anusha

Now, according to ur query

Can case 1) and case 3) be happen in same time?

Means u r saying , only A sold , and A,B both sold ---------these both cases are happening at same time.

Is it possible?

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@ yashgupta1992

Again think

A is only comparable with B.

So, how 2/3 is possible?

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I am also getting 0.789.

My Approach is:-

A = prob of A is sold = 0.6

B = prob of B is sold. = 0.4

A' = prob of A not sold = 0.4

B' = prob of B not sold. = 0.6

We have total four cases as

A'B'

A'B

AB'

AB

Out of these only last three are valid as per question:-

Given that at least one computer has been sold.

So we eliminate 1st case, since no computer sold in that case.

And question is asking Prob of A is sold for this we will have case 3 and case 4, as in these cases Comp A has been sold.

So, plugging the values we have, 

$\frac{0.6 * 0.6 + 0.6* 0.4}{0.6 * 0.6 + 0.6* 0.4 + 0.4*0.4}$

$\frac{60}{76}$

0.789

Please let me know what is wrong in this??