Answer:
(a). 0.0678.
(b). 0.01356
(c). 11.1
(d). 33.3
Explanation:
So, we are given the following data or parameters that is going to help us or assist us in solving this particular problem or Question.
(1). We are given from the question that the archer to hit a bull’s-eye, the probability will be 0.09.
(2). "the results of different attempts can be taken to be independent of each other".
Therefore, let us delve right into the solution to the question.
(a). The probability that the first bull’s-eye is scored with the fourth arrow = P(a= 4) = P × ( 1 - P)^a -1.
P(a = 4) = 0.09 × (1 - 0.09) ^4 - 1.
P(a= 4) = 0.0678.
(b). The probability that the third bull’s-eye is scored with the tenth arrow = P(b= 10, y = 3).
P(b= 10, y = 3) = 36 (0.91)^7 × ( 0.09)^3 = 0.01356.
(c). The expected number of arrows shot before the first bull’s-eye is scored = 1/0.09 = 11.1.
(d). The expected number of arrows shot before the third bull’s-eye is scored = 3/0.09 = 33.3.