Question

According to an​ article, 47​% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly selected​ adults, find the probability that the​ number, X, who have experienced a breakup at least once during the last 10 years is

a. exactly​ five; at most​ five; at least five.

b. at least​ one; at most one.

c. between five and seven​, inclusive.

d. Determine the probability distribution of the random variable X.

Answer 1

a)
Here, n = 9, p = 0.47, (1 - p) = 0.53 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 5)
P(X = 5) = 9C5 * 0.47^5 * 0.53^4
P(X = 5) = 0.228

P(X >= 5) = (9C5 * 0.47^5 * 0.53^4) + (9C6 * 0.47^6 * 0.53^3) + (9C7 * 0.47^7 * 0.53^2) + (9C8 * 0.47^8 * 0.53^1) + (9C9 * 0.47^9 * 0.53^0)
P(X >= 5) = 0.228 + 0.1348 + 0.0512 + 0.0114 + 0.0011
P(X >= 5) = 0.4265

P(X <= 5) = (9C0 * 0.47^0 * 0.53^9) + (9C1 * 0.47^1 * 0.53^8) + (9C2 * 0.47^2 * 0.53^7) + (9C3 * 0.47^3 * 0.53^6) + (9C4 * 0.47^4 * 0.53^5) + (9C5 * 0.47^5 * 0.53^4)
P(X <= 5) = 0.0033 + 0.0263 + 0.0934 + 0.1933 + 0.2571 + 0.228
P(X <= 5) = 0.8014

b)
P(X = 0) = 9C0 * 0.47^0 * 0.53^9
P(X = 0) = 0.0033
P(X>=1) = 1 - 0.0033 = 0.9967

P(X <= 1) = (9C0 * 0.47^0 * 0.53^9) + (9C1 * 0.47^1 * 0.53^8)
P(X <= 1) = 0.0033 + 0.0263
P(X <= 1) = 0.0296

c)
P(5 <= X <= 7) = (9C5 * 0.47^5 * 0.53^4) + (9C6 * 0.47^6 * 0.53^3) + (9C7 * 0.47^7 * 0.53^2)
P(5 <= X <= 7) = 0.228 + 0.1348 + 0.0512
P(5 <= X <= 7) = 0.414

d)
Binomial with p = 0.47 and n = 9