A bag contains 6 tennis balls and 8 cricket balls.Three balls are randomly drawn from the bag.what is the probability that atleast one of the drawn balls is cricket ball
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Let drawing of a cricket ball be treated as success.
And x denote random variable of occurrence of success.
So,we are interested in p(x>=1).
This can be also computed as p(x>=1)=1-p(x=0)
Now probability of drawing a cricket ball =8/14=4/7 (since there are 8 cricket balls in 14 total balls)
The question doesn't specifies that in how many trials atleast 1 cricket ball is drawn.
Let us say total no of trials is n.
This leads to a problem of binomial distribution. Where p=4/7
Now,p(x=0)= nc0 ×(4/7)^0×(3/7)^n
=(3/7)^n
So,required probability=1-(3/7)^n
Where n is total no of trials.
syedmunazza:
options are 2/91, 3/91,5/91 and 86/91 which one is correct
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