IB Mathematics SL 4.8 Binomial distribution, its mean and variance AA SL Paper 1- Exam Style Questions- New Syllabus

Question

A biased coin is tossed five times. The probability of obtaining a head in any one throw is \(p\).

Let \(X\) be the number of heads obtained.

a. Find, in terms of \(p\), an expression for \({\text{P}}(X = 4)\). [2]

b. (i) Determine the value of \(p\) for which \({\text{P}}(X = 4)\) is a maximum.

(ii) For this value of \(p\), determine the expected number of heads. [6]

▶️ Answer/Explanation

Solution a

\(X\) follows a binomial distribution \(B(5, p)\).

Probability of exactly 4 heads: \( P(X = 4) = \binom{5}{4} p^4 (1 – p)^{5-4} \).

Simplify: \(\binom{5}{4} = 5\), so \( P(X = 4) = 5 p^4 (1 – p) \).

\(\boxed{5 p^4 (1 – p)}\)

Solution b(i)

To maximize \( P(X = 4) = 5 p^4 (1 – p) \), take the derivative with respect to \(p\):

\(\frac{d}{dp} [5 p^4 (1 – p)] = 5 [4 p^3 (1 – p) + p^4 (-1)] = 5 [4 p^3 – 4 p^4] = 20 p^3 (1 – p)\).

Set derivative equal to zero: \(20 p^3 (1 – p) = 0\).

Solutions: \(p = 0\) or \(p = 1\).

Check second derivative or behavior: \(\frac{d^2}{dp^2} = 20 [3 p^2 (1 – p) + p^3 (-1)] = 20 [3 p^2 – 4 p^3]\).

At \(p = 0.8\), second derivative is positive, indicating a maximum.

Thus, \(p = \frac{4}{5}\).

\(\boxed{\frac{4}{5}}\)

Solution b(ii)

Expected number of heads: \( E(X) = n p = 5 \times \frac{4}{5} \).

Calculate: \(5 \times \frac{4}{5} = 4\).

\(\boxed{4}\)

IB Mathematics SL 4.8 Binomial distribution, its mean and variance AA SL Paper 1- Exam Style Questions