In a binomial distribution with n = 5 and p = 0.4, what is P(X = 3)?

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Multiple Choice

In a binomial distribution with n = 5 and p = 0.4, what is P(X = 3)?

Explanation:
In a binomial scenario, the chance of exactly k successes in n independent trials is found by combining the ways to choose which trials are successes with the probability of that specific configuration. Here, to have exactly 3 successes out of 5 with a success probability of 0.4: - There are C(5,3) = 10 ways to pick which three trials are successes. - The probability of any particular set of three successes and two failures is (0.4)^3 × (0.6)^2 = 0.064 × 0.36 = 0.02304. Multiply the number of ways by this probability: 10 × 0.02304 = 0.2304. So, P(X = 3) = 0.2304 (about 23.04%).

In a binomial scenario, the chance of exactly k successes in n independent trials is found by combining the ways to choose which trials are successes with the probability of that specific configuration. Here, to have exactly 3 successes out of 5 with a success probability of 0.4:

  • There are C(5,3) = 10 ways to pick which three trials are successes.
  • The probability of any particular set of three successes and two failures is (0.4)^3 × (0.6)^2 = 0.064 × 0.36 = 0.02304.

Multiply the number of ways by this probability: 10 × 0.02304 = 0.2304.

So, P(X = 3) = 0.2304 (about 23.04%).

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