A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.

Options

A. `(25)/(128)`
B. `(13)/(216)`
C. `(25)/(216)`
D. `(11)/(128)`

Answer

C

Explanation

(c ) it is case of Bernoulli trials with n=4. Here, success is getting a doublet.
When a pair of dice is thrown once there are
`6xx6=36` equally likely outcomes.
`rArr` Total number of cases=36
And possible doublets are
`{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}`
`rArr` Favourable number of cases=6
`:. " "p=p("success")=P`
(getting a doublet in a single throw of a pair of dice)
`=("Favorable number of cases")/("Total number of cases")=(6)/(36)=(1)/(6)`
and `" " q=p("failure")=1-p=1-(1)/(6)=(5)/(6)`
Clearly, X has the binomial distribution with `n=4,p=(1)/(6)` and `q=(5)/(6)`.
Now, `" " P(X=r)=""^(n)C_(r )p^(r )q^(n-r)`, where r=0,1,2,.....,n.
`rArr " "P(X-r)=""^(4)C_(r )((1)/(6))^(r ).((5)/(6))^(4-r)=""^(4)C_(r )(5^(4-r))/(6^(4))`
Now, P(2 successes)`=""^(4)C_(2)p^(2)q^(2)=(4xx3)/(2)((1)/(6))^(2)((5)/(6))^(2)`
`=6xx(1)/(36)xx(25)/(36)=(25)/(216)`