The surface of a circular dart board has a smallcenter circle called the bull’s-

Question

The surface of a circular dart board has a smallcenter circle called the bull’s-eye and 20 pie-shaped regions numbered from 1 to 20. Each of the pie-shaped regions is further divided into three parts such that a person throwing a dart that lands in a specific region scores the value of the number, double the number, or triple the number, depending on which of the three parts the dart hits. If a person hits the bull’s-eye with probability 0.01, hits a double with probability 0.10, hits a triple with probability 0.05, and misses the dartboard with probability 0.02, what is the probability that 7 throws will result in no bull’s-eyes, no triples, a double twice, and a complete miss once?

Explanation / Answer

We have:
P(bull's-eye) = 0.01
P(miss) = 0.02
P(triple) = 0.05
P(double) = 0.10
P(single) = 1 - 0.01 - 0.10 - 0.05 - 0.02 = 0.82

Assuming that the 7 throws are independent, we want to know the probability that:
• Exactly 4 of them were singles.
• Exactly 2 of them were doubles.
• Exactly 1 of them was a miss.

Since order doesn't matter, the probability that this occurs is:
[7!/(4!2!1!)] * (0.82)^4 * (0.10)^2 * (0.02)^1 = 0.00949455696

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