Suppose that a box contains r red balls and w white balls. suppose also that balls are drawn from the box one at a time, at random, without replacement. (a)what is the probability that all r red balls will be obtained before any white balls are obtained? (b) what is the probability that all r red balls will be obtained before two white balls are obtained?

Part a)

part b)

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E =

For part a)

Probability that a red ball is drawn in first attempt =

Probability that a red ball is drawn in second attempt=

Probability that a red ball is drawn in third attempt =

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt =

Thus the probability that events occur in succession is

Thus =

Thus our probability becomes

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals

Thus the probability becomes

Thus required probability of case b becomes

=

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

the answer is two. hope this !