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Nov 2023 p51 q5
3105
A red spinner has four sides labelled 1, 2, 3, 4. When the spinner is spun, the score is the number on the side on which it lands. The random variable X denotes this score. The probability distribution table for X is given below.
x
1
2
3
4
P(X = x)
0.28
p
2p
3p
(a) Show that \(p = 0.12\).
A fair blue spinner and a fair green spinner each have four sides labelled 1, 2, 3, 4. All three spinners (red, blue and green) are spun at the same time.
(b) Find the probability that the sum of the three scores is 4 or less.
(c) Find the probability that the product of the three scores is 4 or less given that X is odd.
Solution
(a) The sum of probabilities must equal 1: \(0.28 + p + 2p + 3p = 1\). Simplifying gives \(0.28 + 6p = 1\). Solving for \(p\), we get \(6p = 0.72\), so \(p = 0.12\).
(b) For fair spinners (blue and green), the probability of any score is 0.25. We consider scenarios where the sum of scores is 4 or less:
R = 1, B = 1, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
R = 1, B = 1, G = 2: \(0.28 \times (0.25)^2 = 0.0175\)
R = 1, B = 2, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
R = 2, B = 1, G = 1: \(0.12 \times (0.25)^2 = 0.0075\)
