Harry has three coins: • One coin is biased so that the probability of obtaining a head when it is thrown is 1/5. • The second coin is biased so that the probability of obtaining a head when it is thrown is 1/4. • The third coin is biased so that the probability of obtaining a head when it is thrown is 1/3. Harry throws the three coins. The random variable X is the number of heads that he obtains.
- (a)Draw up the probability distribution table for X.[4]
- (b)Harry has two other coins, each of which is biased so that the probability of obtaining a head when it is thrown is p. He throws all five coins at the same time. The random variable Y is the number of heads that he obtains. Given that P(Y = 0) = 6P(Y = 5), find the value of p.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [B1] Table with correct X values and at least one probability. Values need not be in order, lines may not be drawn, may be vertical, X and P(X) may be omitted. Condone any additional X values if probability stated as 0.
- [B1] P(X=1) or P(X=2) correct and identified, need not be in table, accept unsimplified.
- [B1] Two more correct and identified probabilities, need not be in table, accept unsimplified.
- [B1] 4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. SC B1 for four probabilities summing to 1 placed in a probability distribution table with the correct x values.
- [B1] Either (2/5)(1-p)² (for P(Y=0)) or (1/60)p² (for P(Y=5)), not (2/4)(1-p)² or (1/60)p²
- [M1] Equating and forming a 3 term quadratic equation. Their P(Y=0) = 6 × their P(Y=5).
- [A1] Not dependent on B1. A0 if p = 2 seen and not clearly rejected. SC B1 if p = 2/3 obtained from a correct quadratic with more than three terms. If p = 2 seen and not clearly rejected, SC B0.
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