Find the probability of getting 5 exactly twice in 7 throws of a die.
It is given that a die is thrown 7 times.
Let the number of times of getting 5 in 7 throws of a die be represented by X.
The probability of getting 5 in a single throw of die is,
p = 1 6
So,
q = 1 − p = 1 − 1 6 = 5 6
Then, X has a binomial distribution with n = 7 and p = 1 6 .
The probability of x successes P ( X = x ) is,
P ( X = x ) = C n x q n − x p x = C 7 x ( 5 6 ) 7 − x ( 1 6 ) x
Where x = 0 , 1 , … , n .
The probability of getting 5 exactly twice is,
P ( X = 2 ) = C 7 2 ( 5 6 ) 5 ( 1 6 ) 2 = 21 ( 5 6 ) 5 ( 1 6 ) 2 = 21 ( 1 36 ) ( 5 6 ) 5 = 7 12 ( 5 6 ) 5
Therefore, the probability of getting 5 exactly twice is 7 12 ( 5 6 ) 5 .
It is given that a die is thrown 7 times.
Let the number of times of getting 5 in 7 throws of a die be represented by X.
The probability of getting 5 in a single throw of die is,
So,
Then, X has a binomial distribution with
The probability of
Where
The probability of getting 5 exactly twice is,
Therefore, the probability of getting 5 exactly twice is
