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Probability

Question
CBSEENMA12033808

Find the probability distribution of white balls drawn when 3 balls are drawn one by one without replacement from a bag containing 4 white and 6 red balls.

Solution

Let X denote the random variable ‘number of white balls’. X can take the values 0, 1, 2, 3.
Let p be the probability of getting first white ball and q the probability of not getting first white ball.
therefore space space space space space straight p space space equals space 4 over 10 comma space space space straight q space equals space 1 space minus space straight p space equals space 1 space minus space 4 over 10 space equals space 6 over 10
straight P left parenthesis straight X space equals space 0 right parenthesis space equals space straight q space straight q space straight q space space equals space 6 over 10 cross times 5 over 9 cross times 4 over 8 space equals space 5 over 30
straight P left parenthesis straight X space equals space 1 right parenthesis space equals space straight p space straight q space straight q space plus space straight q space straight p space straight q space space plus space straight q space space straight q space straight p
space space space space space space space space space space space space space space equals space 4 over 10 cross times 6 over 9 cross times 5 over 8 plus 6 over 10 cross times 4 over 9 cross times 5 over 8 plus 6 over 10 cross times 5 over 9 cross times 4 over 8 space equals space 5 over 30 plus 5 over 30 plus 5 over 30 space equals space 15 over 30
straight P left parenthesis straight X space equals space 2 right parenthesis space equals space straight p space straight p space straight q space plus space straight p space straight q space straight p space space plus space straight q space straight p space space straight p
space space space space space space space space space space space space space space space space space equals space 4 over 10 cross times 3 over 9 cross times 6 over 8 plus 4 over 10 cross times 6 over 9 cross times 3 over 8 plus 6 over 10 cross times 4 over 9 cross times 3 over 8 space equals space 3 over 30 plus 3 over 30 plus 3 over 30 space equals 9 over 30
straight P left parenthesis straight X space equals space 3 right parenthesis space equals space straight p space straight p space straight p space equals space 4 over 10 cross times 3 over 9 cross times 2 over 8 space equals space 1 over 30
∴     probability distribution is

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