Answer:
0.7455
Step-by-step explanation:
First, we note that there are C[tex]C_{12|3}[/tex] = 220 ways of selecting the defective items.
The probability that the inspector will have to test at least 9 widgets would be 1 minus the probability that the inspector will have to test 8 or less widgets.
This is 1 - P(8 or less widgets have to be checked)
1 - [tex]\frac{C_{8|3} }{C_{12|3} }[/tex]
= 1 - 56/220
= 1 - 0.2545
= 0.7455
Answer:
[tex]P[x\geq 9] = 0.0003971[/tex]
Step-by-step explanation:
Given data:
n = 12
Defective widget probability is [tex]p = \frac{3}{12} = 0.25[/tex]
q = 1 - p = 0.75
Probability of test atleast 9 widget is calculated by binomial distribution as
[tex]P[x\geq 9] = P(9) + P(10) +P(11) + P(12)[/tex]
[tex]=\sum_{X =9}^{12} ^12C_x p^x q^{12-x}[/tex]
[tex]= \sum_{X =9}^{12} ^12C_x (0.25)^x (0.75)^{12-x} [/tex]
[tex]= \sum_{X =9}^{12} ^{12}C_9 (0.25)^9 (0.75)^{3} +\sum{x=9}{12} ^{12}C_{10} (0.25)^{10} (0.75)^{2} + \sum{x=9}{12} ^{12}C_{11} (0.25)^{11} (0.75)^{1} +\sum{x=9}{12} ^{12}C_{12} (0.25)^{12} (0.75)^{0}[/tex]
= 0.003541 + 0.000354+ 0.0000021 + 0.0000001
= 0.0003971
[tex]P[x\geq 9] = 0.0003971[/tex]
