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
