Question

In the old days, there was a probability of 80% of success in any attempt to make a telephone call. (This often depended on the importance of the person making the call, or the operator’s curiosity.) In six attempted calls, (a) Construct a probability distribution for X = the number of successful calls in six attempts. (b) Find the probability that at least two calls will be made successfully. (c) Find the mean and standard deviation of the distribution.

Answer 1

a)

x   formula   value
0   6C0 * 0.8^0 * 0.2^6   0.0001
1   6C1 * 0.8^1 * 0.2^5   0.0015
2   6C2 * 0.8^2 * 0.2^4   0.0154
3   6C3 * 0.8^3 * 0.2^3   0.0819
4   6C4 * 0.8^4 * 0.2^2   0.2458
5   6C5 * 0.8^5 * 0.2^1   0.3932
6   6C6 * 0.8^6 * 0.2^0   0.2621

b)

Here, n = 6, p = 0.8, (1 - p) = 0.2 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 2).
P(X >= 2) = (6C2 * 0.8^2 * 0.2^4) + (6C3 * 0.8^3 * 0.2^3) + (6C4 * 0.8^4 * 0.2^2) + (6C5 * 0.8^5 * 0.2^1) + (6C6 * 0.8^6 * 0.2^0)
P(X >= 2) = 0.0154 + 0.0819 + 0.2458 + 0.3932 + 0.2621
P(X >= 2) = 0.9984

c)

Here, μ = n*p = 4.8, σ = sqrt(np(1-p)) = 0.9798