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N=15, p=.45, q=.55 find the probability of fewer than 3

User Holybull
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Answer:

The binomial distribution formula is given by:

P(X = k) = (nCk) * p^k * q^(n-k)

where:

P(X = k) is the probability of getting exactly k successes

n is the number of trials

k is the number of successes

p is the probability of success in a single trial

q is the probability of failure in a single trial (1 - p)

(nCk) is the binomial coefficient, calculated as n! / (k!(n-k)!)

In this case, we have n = 15, p = 0.45, and q = 0.55.

Now, let's calculate the probability of getting fewer than 3 successes (0, 1, or 2).

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = 0) = (15C0) * (0.45^0) * (0.55^15)

P(X = 1) = (15C1) * (0.45^1) * (0.55^14)

P(X = 2) = (15C2) * (0.45^2) * (0.55^13)

Using the binomial coefficient formula (nCk) = n! / (k!(n-k)!), we can calculate:

P(X = 0) = (1) * (1) * (0.55^15)

P(X = 1) = (15) * (0.45) * (0.55^14)

P(X = 2) = (15C2) * (0.45^2) * (0.55^13)

By substituting the given values and evaluating these equations, we can find the probability of fewer than 3 successes.

Explanation:

User TheGeorgeous
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