A bag contains four tickets numbered 00, 01, 10 and 11. Four tickets are chosen at random with replacement, then the probability that sum of the numbers on the tickets is 23, 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)`