Question

Consider a binomial probability distribution with p = 0.3 and n = 9. What is the probability of the​ following? ​

Answer 1

Answer: a) 0.2668, b) 0.4224, c) 0.004

Step-by-step explanation: this is a question under binomial probability.

P(x=r) = nCr * p^r * q^n-r.

Question a)

n = 9, p = 0.3, q = 1 - p = 1 - 0.3 = 0.7

for our question, r =3 ( exactly 3 successes)

P(x=3) = 9C3 * (0.3)^3 * (0.7)^6

P(x=3) = 85 * 0.027 * 0.117649

P(x=3) = 0.2668.

Question b

Probability less that 3 success = p(x<3) = p(x=2) + p(x=1).

At p(x=2)

p(x=2) = 9C2 * (0.3)^2 * (0.7)^7

p(x=2) = 36 * 0.09 * 0.0823543

p(x=2) = 0.2668.

At p(x=1)

p(x=1) = 9C1 * (0.3) * (0.7)^8

p(x=1) = 9 * 0.3 * 0.0576

p(x=1) = 0.1556

p(x<3) = p(x=2) + p(x=1).

p(x<3) = 0.2668 + 0.1556

p(x<3) = 0.4224.

Question c)

P (x≥7) = 1 - p(x≤6)

p(x≤6) can be gotten by using the cumulative binomial table.

From the table, we had that p(x≤6) = 0.996

P (x≥7) = 1 - 0.996

P (x≥7) = 0.004