Eli has four fair 4-sided dice with sides labelled 1, 2, 3, 4. He throws all four dice at the same time. The random variable X denotes the number of 2s obtained.
- (a)Show that P(X = 3) = 3/64.[2]
- (b)Complete the following probability distribution table for X. X | 0 | 1 | 2 | 3 | 4 P(X = x) | 81/256 | | | 3/64 | 1/256[2]
- (c)Find E(X).[2]
- (d)Eli throws the four dice at the same time on 96 occasions. Use an approximation to find the probability that he obtains at least two 2s on fewer than 20 of these occasions.[5]
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Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [M1] (3/4)³ x (1/4)¹ x 4
- [A1] AG.
- [B1] Either P(1) = 27/64, 0.421875 or P(2) = 27/128, 0.2109375 correct to at least 3SF. Condone not in table.
- [B1 FT] Both values in table. FT P(1) + P(2) = 81/128, 0.6328125.
- [M1] Correct method from their probability distribution table with at least 4 terms, 0< their P(x) < 1, accept partially evaluated.
- [A1]
- [B1] 25.125, 25 1/8 and 18.5493... to at least 3SF seen, allow unsimplified (4.3068 ≤ σ ≤ 4.307 implies correct variance).
- [M1] Substituting their μ and σ into ± standardisation formula (any number for 19.5). Condone σ² and √σ.
- [M1] Using continuity correction 19.5 or 20.5 in their standardisation formula. Note: ±5.625/√18.549 seen gains M2 BOD.
- [M1] Appropriate area Φ, from final process. Must be a probability.
- [A1] 0.0957 < p < 0.0958 . SC B1 for 0.0957 < p < 0.0958 if B1M0M0M1 scored.
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