From the given data probability of malfunction is given by,
"P(p)=0.007\\\\\nP(q)=0.003\\\\\nP(r)=0.001\\\\"
And for each of the machines , probability of not malfunctioning is given by,
"P'(p)=1-0.007=0.993\\\\\nP'(q)=1-0.003=0.997\\\\\nP'(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)\\\\\nP(E)=0.993\\times0.997\\times0.001+0.993\\times0.003\\times0.999+0.007\\times0.997\\times0.999\\\\\nP(E)=0.00099+0.00297+0.00697\\\\\nP(E)=0.0109"
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