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Random Variables, Expectation & Variance · Question 29 of 68

Question 29

HardOriginal question

XX takes the values 0 to 7 with probabilities 0,k,2k,2k,3k,k2,2k2,7k2+k0, k, 2k, 2k, 3k, k^{2}, 2k^{2}, 7k^{2} + k. What is kk?

Options for question 29

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. The probabilities add up to 1: 10k2+9k=110k^{2} + 9k = 1, that is (10k−1)(k+1)=0(10k - 1)(k + 1) = 0.

  2. A probability cannot be negative, so k=110k = \frac{1}{10}, not −1-1.

Worth noting

19\frac{1}{9} drops the k2k^{2} terms and solves 9k=19k = 1.

Concept
Finding the constant in a distribution
Typical time
120 s
Your time
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The rule this testsProbability distributions
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Ten questions from Random Variables, Expectation & Variance, at your own pace.

Practise this chapter →

Ten questions from Random Variables, Expectation & Variance, at your own pace.