Alisha has four coins. One of these coins is biased so that the probability of obtaining a head is 0.6. The other three coins are fair. Alisha throws the four coins at the same time. The random variable X denotes the number of heads obtained.
- (a)Show that the probability of obtaining exactly one head is 0.225.[3]
- (b)Complete the following probability distribution table for X. X 0 1 2 3 4 P(X = x) 0.05 0.225 0.075[2]
- (c)Given that E(X) = 2.1, find the value of Var(X).[2]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.
- [B1] Either 0.6(0.5)³ + a or b + 0.4(0.5)³×(3or 3C₁), 0 < a, b < 1 seen.
- [M1] 0.6(0.5)³ +0.4(0.5)³×d seen, d = 1, 3. Condone 0.075 +0.05 × d, d = 1, 3.
- [A1] AG full supporting working required. Scenarios identified and linked to calculations.
- [B1] Either [P(2) =] 0.375, 3/8 or [P(3) =] 0.275, 11/40 seen. Condone not in table if identified.
- [B1 FT] Both values in table. FT P(2) + P(3) = 0.650 .
- [M1] Appropriate variance formula from their probability distribution table with at least 4 terms, 0 < their P(x) < 1. Condone 4.41 for 2.1². Condone mean clearly recalculated inaccurately. Or 0.225+4×their0.375+9×their0.275+16×0.075-2.1². Condone 2.1² for 4.41.
- [A1] [5.4-2.1²]=0.99[0]. If M0 awarded SC B1 for 0.99[0] WWW.
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