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Probability

Question
CBSEENMA12033812

Find the probability distribution of number of doublets in three throws of a pair of dice.

Solution

 Let X denote the number of doublets. X can take the value 0, 1, 2, or 3.
Possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Probability of getting a doublet  = 6 over 36 space equals 1 over 6
Probability of not getting a doublet  = 1 minus 1 over 6 space equals 5 over 6
Now,   P(X = 0) = P(no doublet) = 5 over 6 cross times 5 over 6 cross times 5 over 6 space equals space 125 over 216
           straight P left parenthesis straight X space equals space 1 right parenthesis space equals space straight P left parenthesis one space doublet space and space two space non minus doublets right parenthesis
space space space space space space space space space space space space space space space space space space equals space 1 over 6 cross times 5 over 6 cross times 5 over 6 plus 5 over 6 cross times 1 over 6 cross times 5 over 6 plus 5 over 6 cross times 5 over 6 cross times 1 over 6
space space space space space space space space space space space space space space space space space space space equals 25 over 216 plus 25 over 216 plus 25 over 216 space equals 75 over 216
           straight P left parenthesis straight X space equals space 2 right parenthesis space equals space straight P left parenthesis two space doublets space and space one space non minus doublet right parenthesis
space space space space space space space space space space space space space space space space space equals space 1 over 6 cross times 1 over 6 cross times 5 over 6 plus 1 over 6 cross times 5 over 6 cross times 1 over 6 plus space 5 over 6 cross times 1 over 6 cross times 1 over 6
space space space space space space space space space space space space space space space space space equals space 5 over 216 plus 5 over 216 plus 5 over 216 space equals space 15 over 216
and straight P left parenthesis straight X space equals space 3 right parenthesis space equals space straight P left parenthesis three space doublets right parenthesis
                         equals space 1 over 6 cross times 1 over 6 cross times 1 over 6 space equals space 1 over 216
∴ the required probability distribution is

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