Question
Jerry makes handcrafted chocolates. On average, 1 in 25 of the chocolates that Jerry makes is flawed. Whether or not a chocolate is flawed is independent of all other chocolates.
(a) In a batch of 20 chocolates, chosen at random, find the probability that:
(i) Two are flawed.
(ii) More than two are flawed.
Jerry sells the perfect chocolates for 50 pesos each and the flawed ones for 15 pesos each.
(b) Calculate the expected number of pesos Jerry makes from selling a batch of 20 randomly selected chocolates.
▶️ Answer/Explanation
Detailed Solution
(a) Recognizing the Binomial Distribution
The number of flawed chocolates follows a binomial distribution:
\[ M \sim B(20, 0.04) \]
where:
- \( n = 20 \) (total chocolates)
- \( p = 0.04 \) (probability of a chocolate being flawed)
(i) Finding the Probability That Exactly Two Chocolates Are Flawed
Using the binomial probability formula:
\[ P(M=2) = \text{binpdf}(20, 0.04, 2) \]
\[ = \binom{20}{2} (0.04)^2 (0.96)^{18} \]
\( P(M=2) = 0.146 \) (or 0.145799…)
(ii) Finding the Probability That More Than Two Chocolates Are Flawed
\[ P(M \geq 3) = 1 – P(M \leq 2) \]
Using cumulative binomial probability:
\[ P(M \geq 3) = 1 – \text{bincdf}(20, 0.04, 2) \]
\( = 0.0439 \) (or 0.0438627…)
(b) Expected Value of Earnings
The expected number of flawed chocolates in 20 is:
\[ E(M) = 20 \times 0.04 = 0.8 \]
The expected number of perfect chocolates is:
\[ E(P) = 20 – 0.8 = 19.2 \]
The expected revenue is:
\[ \text{Revenue} = 50(19.2) + 15(0.8) \]
\[ = 960 + 12 \]
\( = 972 \) pesos
……………………………Markscheme……………………………….
(a) (i) Probability of exactly two flawed chocolates: 0.146
(a) (ii) Probability of more than two flawed chocolates: 0.0439
(b) Expected revenue: 972 pesos