Question

A marketing manager makes the statement that the long-run probability that a
customer would prefer the deluxe model to the standard model is 30%.
1.2.1 What is the probability that exactly 3 in a random sample of 10 customers will
prefer the deluxe model? (3)
1.2.2 What is the probability that more than 2 in a random sample of 10 customers
will refer the standard model?

Answer 1

Answer on Question#42679 – Math – Statistics and Probability

Question:

A marketing manager makes the statement that the long-run probability that a customer would prefer the deluxe model to the standard model is 30%.

1.2.1 What is the probability that exactly 3 in a random sample of 10 customers will prefer the deluxe model? (3)

1.2.2 What is the probability that more than 2 in a random sample of 10 customers will refer the standard model?

Solution.

The probability that a customer would prefer the deluxe model to the standard model is $p=0.3$.

We will use the Bernoulli scheme:

$$P(x = k) = C_n^k p^k (1 - p)^{n - k}$$

1.2.1 The probability that exactly 3 in a random sample of 10 customers will prefer the deluxe model is

$$P(x = 3) = C_{10}^3 (0.3)^3 (0.7)^7 = \frac{10!}{3!7!} * (0.3)^3 (0.7)^7 = 0.266827932$$

1.2.2 The probability that more than 2 in a random sample of 10 customers will prefer the standard model is $P(x > 2) = 1 - P(x \leq 2) = 1 - P(x = 1) - P(x = 2) - P(x = 0)$.

The probability that a customer would prefer the standard model is $p=0.7$

$$P(x = 0) = C_{10}^0 0.7^0 (0.3)^{10} = 1 * 1 * (0.3)^{10} = 0.0000059049$$ $$P(x = 1) = C_{10}^1 0.7 (0.3)^9 = 10 * 0.7 (0.3)^9 = 0.000137781$$ $$P(x = 2) = C_{10}^2 (0.7)^2 (0.3)^8 = \frac{10!}{2! * 8!} (0.7)^2 (0.3)^8 = 0.0014467005$$ $$\begin{array}{l} P(x > 2) = 1 - P(x \leq 2) = 1 - P(x = 1) - P(x = 2) \\ = 1 - 0,000137781 - 0,0014467005 - 0,0000059049 = 0,9984096136 \\ \end{array}$$

Answer. 0,9984096136.

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