Probability problem

ANSWER - 1

There seems some errors in forming question.According to me, question should be read as follow :-

1. given 30% is the passing probability for physics

2. given 40% is the passing probability for chemistry

3. given 50% is the probability of failed in atleast in One subject out of two i.e. either in 1+ in both subjects

Let there be 100 students

50% prob. of failed in atleast in one = prob. of pass physics + prob. of pass chemistry - prob. of passing both subjects

50 = 30 + 40 - n(Passed in both Phy & Chem)

n(Passed in both Phy & Chem) = 20.

Ppl who passed in ONLY Physics = 30 - 20 = 10

If the person who got selected did fail in chemistry, there are 60 such people.

P(Passed in Physics given that the person failed in Chem) = 10 / 60 = 1/6.

ANSWER - 2

We have to find prob. of not more than 1 defective i.e. in sampling, we may get 0 or 1 defective

Required Prob. = 1 - prob. of getting 2 defectives

                             = 1 - (2C2 * 8C2 / 10C4)

                             = 1 - 2/15

                             = 13/15

                                                                     OR

Required Prob. = prob of getting 0 defective + prob of getting 1 defective

                             = ((2C0 * 8C4) + (2C1 * 8C3)) / 10C4

                             = 13/15

ANSWER - 3

1. The first ball can be distributed to 1st box or 2nd box or 3rd box i.e. it can be distributed in 3 ways

2. Similarly 2nd ball also can be distributed in 3 ways. Thus First 2 ball can be distributed in 3power2

3. In this way, 8 balls can be distributed to 3 boxes in 3power8

Now there is requirement that first box should contain 3 balls which implies that remaining 5 balls

must go to remaining 2 boxes and in 2power5 (as above in  1 & 2)           

3 balls out of 8 balls can be selected in 8C3 ways, Thus

Required Prob. = (8C3 * 2power5) / 3power8

                            = (56*32) / 6561

                           = 1792 / 6561

                           = 0.2731