A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses 3 sweets at random from a box.
- (a)Draw up the probability distribution table for the number of jellies that Jemeel chooses.[4]
- (b)The company also produces large boxes of sweets. For any large box, the probability that it contains more jellies than chocolates is 0.64. 10 large boxes are chosen at random. Find the probability that no more than 7 of these boxes contain more jellies than chocolates.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [1] B1 for probability distribution table with correct outcome values (X: 0, 1, 2, 3)
- [1] M1 for denominator 8C3 or 8×7×6
- [1] Any one probability correct (with correct outcome)
- [1] All probabilities correct (P(0)=1/56, P(1)=15/56, P(2)=30/56, P(3)=10/56)
- [1] 1-P(8, 9, 10) = 1 - [10C8 0.64^8 0.36^2 + 10C9 0.64^9 0.36^1 + 0.64^10]
- [1] 1-(0.164156+0.064852+0.011529)
- [1] 0.759
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