`Aa n dB` toss a fair coin each simultaneously 50 times. The probability that both of them will not get tail at the same toss is `(3//4)^(50)` b. `(2//7)^(50)` c. `(1//8)^(50)` d. `(7//8)^(50)`
A. `(3//4)^(50)`
B. `(2//7)^(50)`
C. `(1//8)^(50)`
D. `(7//8)^(50)`

10 views

1 Answers

Correct Answer - A
For each toss, there are four choices:
(i) A gets head, B gets head
(ii) A gets tail, B gets head
(iii) A gets head, B gets tail
(iv) A gets tail, B gets tail
Thus, exhaustive number of ways is `4^(50)`. Out of the four choices listed above, (iv) is not favorable to the required event in a toss. Therefore, favorable number of cases is `3^(50)`. Hence, the required probability is `(3//4)^(50)`.

10 views

Related Questions