Question

Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine if the events are unusual. If convenient, use the appropriate probability table or technology to find the probabilities. A football player completes a pass 67.3% of the time. Find the probability that (a) the first pass he completes is the second pass, (b) the first pass he completes is the first or second pass, and (c) he does not complete his first two passes. (a) P(the first pass ho completes is the second pass) - (Round to three decimal places as needed.) (b) P(the first pass he completes is the first or second pass)- (Round to three decimal places as needed) (c) P(he does not complete his first two passes)- (Round to three decimal places as needed.)

Answer 1

example:

DEFINITIONS Definition binomial probability: p\.(1 - 2)^-^ n! P(X = k) =. Ckpk. (1 – p)n-k k!(n - k)! Definition geometric probability: P(2) = 7"-1p= (1 - p)-1 Formula Poisson probability: "e- P(x) = Addition rule for disjoint or mutually exclusive events: P(AUB) = P(A or B) = P(A) + P(B) Complement rule: P(A) = P( not A) = 1 - P(A)

SOLUTION p = 63.7% = 0.637 The binomial distribution needs to be used when we are interested in the number of successes among a fixed number of trials. The geometric distribution needs to be used when we are interested in the occurrence of the first success. The Poisson distribution needs to be used when we are interested in the number of successes over a certain (time) interval. In this case, we are interested in the first success (sale) and thus we need to use the geometric distribution. (a) Evaluate the definition of geometric probability at x = 2: P(2) = (1 -0.637)2-1 x 0.637 = 0.3631 x 0.637 = 0.363 x 0.637 = 0.231231

(b) Evaluate the definition of geometric probability at r = 1,2 P(1) = (1 -0.637)1-1 x 0.637 = 0.3630 x 0.637 = 1 x 0.637 = 0.637 P(2) = (1 -0.637)2-1 x 0.637 = 0.3631 x 0.637 = 0.363 x 0.637 = 0.231231 Use the addition rule mutually exclusive events: P(1 or 2) = P( X2) = P(1) + P(2) = 0.637+0.231231 = 0.868231

(c) Use the complement rule: P(X > 2) = 1 - P( X2) =1-0.868231 = 0.131769

Criven, p = 67.3% = 0.673 12-1 X 0.673 2. P(9) - (1 - 0-673) (0:329) * x 0-673 Geometrie probability 0.220071 x 0.673 ::P(1) - (1 - 0.673) (6.3273 x 0-673 Geometric Probability 1 X 0.673 = 0.673 ...P(B) = (2-0673) 2-1 X0-673 = 0.9.2007 ..P(2012) - P(x + 2) - P(1) + P(3) mutual exclusie = 0.673 +0.220071 = 0.893071 ① Use the complement rule: P(x-2) = = 1- P(x22) -1- 0.893071 0.106929. So, we can say that mone of the events unusual