Answer :

Given that, obtaining multiple of 3 in a throw of a die is a success.

There are total 10 throws of a fair die.

Let p be the probability of getting multiple of 3 in a throw of a die.

Since there can be a total 6 outcomes in a throw of a die. That is, {1, 2, 3, 4, 5, 6}.

And a multiple of 3 in a die is {3}, {6}.

Then, let q be the probability of not getting a multiple of 3 in a throw of a die.

And we know, p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of successes (getting multiple of 3 in die) out of 10 throws of a die.

Then, the probability of getting r successes out of n throws of die is given by,

P (X = r) = ^{n}C_{r}p^{r}q^{n-r}

Here n = 10.

Now, put values of n, p, and q in the above equation.

…(i)

We need to find the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die.

It is given,

Probability = P (X ≥ 8)

This can be written as,

P (X ≥ 8) = P (X = 8) + P (X = 9) + P (X = 10)

Just out r = 8, 9, 10 in equation (i) to find the value of P (X = 8), P (X = 9), P (X = 10) respectively, then substitute in the above equation.

Thus, the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die is 201/3^{10}.

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