****I already have the answers and formulas for this, i need steps to see how to
ID: 3171740 • Letter: #
Question
****I already have the answers and formulas for this, i need steps to see how to enter it in my calculator please!****
A company installs new central heating furnaces, and has found that for 15% of all installations a return visit is needed to make some modifications. Six installations were made in a particular week. Assume independence of outcomes for these installations.
a. What is the probability that a return visit was needed in all of these cases?
b. What is the probability that a return visit was needed in none of these cases?
c. What is the probability that a return visit was needed in more than one of these cases?
Explanation / Answer
p = 0.15
n = 6
P(X = x) = 6Cx * 0.15x * (1 - 0.15)6-x
a) P(X = 6) = 6C6 * 0.156 * 0.850
= 1.139 * 10-5
b) P(X = 0) = 6C0 * 0.150 * 0.856
= 0.3771
c) P(X > 1) = 1 - P(X < 1)
= 1 - [P(X = 0) + P(X = 1)]
= 1 - [6C0 * 0.150 * 0.856 + 6C1 * 0.151 * 0.855]
= 1 - 0.7765
= 0.2235
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