What is the most probable number of successes in 10 trials with probability of success `2/3`? (d) 4 (b) 7 (a) 10 (c) 5

Options

A. 10
B. 7
C. 5
D. 4

Answer

B

Explanation

This is the question of Binomial Distribution.
Number of r sucesses in n trial is
`P_(r)=.^(n)C_(r) p^(r) q^(n-r)`
When p=prob of success
q=prob, of failure=1-p
Given n-10, `p=(2)/(3),q=(1)/(3)`
`Fo r=10, P_(10)=.^(10)C_(10) ((2)/(3))^(10)((1)/(3))^(0)=(2^(10))/(3^(10))`
`"For" r=7, P_(7)=120. (2^(7))/(3^(10))`
`"For " r=5, P_(5)=252xx(2^(5))/(3^(10)`
`"For "r=4, P_(4)=210.(2^(4))/(3^(10))`