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Ira Lisetskai [31]
2 years ago
6

A widget inspector inspects 12 widgets and finds that exactly 3 are defective.Unfortunately, the widgets then get all mixed up a

nd the inspector has to find the 3defective widgets again by testing widgets one by one.(a) Find the probability that the inspector will now have to test at least 9 widgets.
Mathematics
2 answers:
WITCHER [35]2 years ago
8 0

Answer:

0.7455

Step-by-step explanation:

First, we note that there are CC_{12|3} = 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 - \frac{C_{8|3} }{C_{12|3} }

= 1 - 56/220

= 1 - 0.2545

= 0.7455

rusak2 [61]2 years ago
6 0

Answer:

P[x\geq 9] = 0.0003971

Step-by-step explanation:

Given data:

n =  12

Defective widget probability is p = \frac{3}{12} = 0.25

q = 1 - p = 0.75

Probability of test atleast 9 widget is calculated by binomial distribution as

P[x\geq 9] = P(9) + P(10) +P(11) + P(12)

=\sum_{X =9}^{12} ^12C_x p^x q^{12-x}

= \sum_{X =9}^{12} ^12C_x (0.25)^x (0.75)^{12-x}

= \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}

= 0.003541 + 0.000354+ 0.0000021 + 0.0000001

= 0.0003971

P[x\geq 9] = 0.0003971

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