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Probability

Question
CBSEENMA12033841
Wired Faculty App

A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
i) none is white?    (ii) all are white?  (iii) only 2 are white?    (iv) at least one is white?

Solution

Here,  n = 4,  p = fraction numerator 5 over denominator 5 plus 7 plus 8 end fraction space equals 1 fourth
             straight q space equals space 1 minus straight p space equals space 1 minus 1 fourth space equals space 3 over 4
(i) P(none is white) = P(0) = straight C presuperscript 4 subscript 0 space open parentheses 3 over 4 close parentheses to the power of 4 minus 0 end exponent space open parentheses 1 fourth close parentheses to the power of 0 space equals space 1 cross times open parentheses 3 over 4 close parentheses to the power of 4 space cross times space 1 space equals space 81 over 256
(ii) P(all white) = P(4) = straight C presuperscript 4 subscript 4 space open parentheses 3 over 4 close parentheses to the power of 4 minus 4 end exponent space open parentheses 1 fourth close parentheses to the power of 4 space equals space 1 cross times 1 cross times 1 over 256 space equals 1 over 256
(iii) P(only 2 are white) = P(2) = straight C presuperscript 4 subscript 2 open parentheses 3 over 4 close parentheses to the power of 4 minus 2 end exponent space open parentheses 1 fourth close parentheses squared space equals space fraction numerator 4 cross times 3 over denominator 1 cross times 2 end fraction cross times 9 over 16 cross times 1 over 6 space equals 27 over 128
(iv) P(at least one is white) = P(1) + P(2) + P(3) + P(4)
                                             equals space 1 minus straight P left parenthesis 0 right parenthesis space equals space 1 minus 81 over 256 space equals fraction numerator 256 minus 81 over denominator 256 end fraction space equals 175 over 256

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