From the given data probability of malfunction is given by,
P(p)=0.007P(q)=0.003P(r)=0.001
And for each of the machines , probability of not malfunctioning is given by,
P′(p)=1−0.007=0.993P′(q)=1−0.003=0.997P′(r)=1−0.001=0.999
Since malfunctions are independent,
Probability that exact one of the dispensers malfunctions is given by,
P(E)=P′(p)P′(q)P(r)+P′(p)P(q)P′(r)+P(p)P′(q)P′(r)P(E)=0.993×0.997×0.001+0.993×0.003×0.999+0.007×0.997×0.999P(E)=0.00099+0.00297+0.00697P(E)=0.0109
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