If a random variable X is binomially distributed with parameters n=5, p=0.4, then the probability of three successes is:a.0.18b.0.23c.0.36d.0.68

The probability of getting exactly k successes in n trials is given by the formula for the Binomial distribution:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:

• C(n, k) is the number of combinations of n items taken k at a time,
• p is the probability of success on a single trial,
• n is the number of trials,
• k is the number of successes we want.

In this case, we have n=5 (the number of trials), p=0.4 (the probability of success on a single trial), and we want to find the probability of k=3 successes.

So, we have:

P(X=3) = C(5, 3) * (0.4^3) * ((1-0.4)^(5-3))

Calculating the combinations, C(5, 3) = 5! / [3!(5-3)!] = 10.

So,

P(X=3) = 10 * (0.4^3) * (0.6^2) = 10 * 0.064 * 0.36 = 0.2304

So, the probability of getting exactly 3 successes is approximately 0.23. Therefore, the correct answer is b. 0.23.