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Suppose X has a binomial distribution B(...

Suppose X has a binomial distribution `B(6,1/2)`. Show that `X = 3`is the most likely outcome.(Hint: `P(x=3)`is the maximum among all `P(x_i),x_i=0,1,2,3,4,5,6)`

A

X=1

B

X=3

C

X=0

D

X=6

Text Solution

Verified by Experts

The correct Answer is:
B
We have , n=6 and p`=1/2` and q`=1-p=1-1/2=1/2`
Then `, P(X=r)=""^(8)C_(r).((1)/(2))^(r)((1)/(2))^(6-r)`
Here , P(X=0) `=""_(6)C_(0)p^(0)q^(6)=""^(6)C_(0)((1)/(2))^(6)=(1)/(2^(6))=(1)/(64)`
P(X=1) `=""^(6)C_(1)p^(1)q^(5) =6(1)/(2)(1/2)^(5)=6(1)/(2^(6))=(1)/(64)`
`P(X=1) =""^(6)C_(1)p^(1)q^(5)=6(1/2)(1/2)^(5)=6((1)/(2^(6)))=(6)/(64)`
`P(X=2) =""^(6)C_(2)P^(2)q^(4)=(6xx5)/(1xx2)*(1/2)^(2)(1/2)^(4)=(15)/(64)`
`P(X=3)=""^(6)C_(3)p^(3)q^(3)=(6xx5xx4)/(1xx2xx3)(1/2)^(3) (1/2)^(3)=(20)/(64)`
`P(X=4)=""^(6)C_(4)p^(4)q^(2)=""^(6)C_(2)(1/2)^(4)(1/2)^(2)=(15)/(64)`
`P(X=5)=""^(6)C_(5)p^(5)q^(1)=""^(6)C_(1)(1/2)^(5)(1/2)^(1)=(6)(64)`
and `P(X=6)=""^(6)C_(6)p^(6)q^(0)=1xx((1)/(2))^(6)=(1)/(64)`
It is clear that `(20)/(64)` is maximum of all the above values . This means that X=3 is the most likely outcome.
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