Four tickets are marked 00, 01, 10 and 11 respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the five tickets drawn is 24, is

Options

A. `(25)/(256)`
B. `(25)/(512)`
C. `(25)/(1024)`
D. `(25)/(128)`

Answer

B

Explanation

We have,
Number of ways of drawing five tickets `=4xx4xx4xx4xx4=4^(5)`
So, total number of elementary events `=4^(5)`
Now,
Favourable number of elementary events
= Number of ways of getting 24 as the sum of the numbers on the five tickets.
= Coefficient of `x^(24) " in " (x^(0)+x^(1)+x^(10)+x^(11))^(5)`
= Coefficient of `x^(24) " in " (1+x)^(5)(1+x^(10))^(5)`
= Coefficient of `x^(24) " in " (1+x)^(5) (.^(5)C_(0)+ .^(5)C_(1) x^(10)+ .^(5)C_(2)x^(20) x^(20) +..)`
`= .^(5)C_(2)xx " Coefficient of " x^(4) " in" (1+x)^(5)= .^(5)C_(2)xx .^(5)C_(4)=50`
Hence, required probability `=(50)/(4^(5))=(25)/(512)`