therealandrew

Will we see Gizmo on the big screen again any time soon?

This is a binomial distribution

This is the reverse of the usual sort of question, in which you are given the individual probability, and have to calculate the multiple probabilities.

If the probability of a single person having a red ticket is p, then the probability of not having a red ticket is (1 - p), which I shall call q .

The sample size is 10, so the probabilities of 0, 1, 2, . . . . . 9, 10 people in the sample having the red ticket is given by the terms of the expansion of

(q + p)¹⁰ = 10C0 (q¹⁰⁾ + 10C1 (q⁹⁾ (p) + 10C2 (q⁸⁾ (p²⁾ +

. . . . . . . . . . . . . . . . . . . . 10C3 (q⁷⁾ (p⁶⁾ + 10C4 (q⁶⁾ (p⁴⁾ + . . . . . etc

which is as far as we need to go, because the term 10C4 (q⁶⁾ (p⁴⁾ is the one we are interested in here.

We are told that 10C4 (q⁶⁾ (p⁴⁾ = 0.2508

The easiest way to find the values of q and p is to refer to published tables of the binomial distribution for n = 10. Look in the body of the table for a value of 0.2508 or thereabouts.

Fortunately, the exact value 0.2508 appears for a value of q = 0.6, p = 0.4

Which means that the probability of any single person having a red ticket is 0.4

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The Probability of exactly 4 people out of 10 having a red ticket at a charity ball is approximately 0.2508. What is the probability, as a percentage, of a random person having a red ticket? (Hint: red or not red).