Question

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.

Answer 1

Here success is a score which is a multiple of 3

i.e., 3 or 6.

Therefore, P(3 or 6) = `2/6 = 1/3`

The probability of r successes in 10 throws is given by

P(r) = `""^10"C"_"r" (1/3)^"r" (2/3)^(10 - "r")`

Now P(At least 8 successes) = P(8) + P(9) + P(10)

= `""^10"C"_8 (1/3)^8 (2/3)^2 + ""^10"C"_9 (1/3)^9 (2/3)^1 + ""^10"C"_10 (1/3)^10`

= `1/3^10 [45 xx 4 + 10 xx 2 + 1]`

= `201/3^10`.