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1)A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
2)Suppose the bulb drawn in (1) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
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total undefective buld =20-4=16 but pe of defective one== 4/20=1/5
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Heya!
1) Total number of outcomes = Total number of bulbs = 20
No. of favourable outcomes = No. of defective bulbs = 4
P(drawing a defective bulb) = 4/20
= 1/5
====
2) Total number of outcomes = 20-1 ( 1 --> a not defective bulb was taken away in first trial and is not replaced. ) = 19
No of favourable outcomes = 19 - 4(defective) = 15
P(drawing a bulb which is not defective) = 15/19
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HOPE IT HELPS!
1) Total number of outcomes = Total number of bulbs = 20
No. of favourable outcomes = No. of defective bulbs = 4
P(drawing a defective bulb) = 4/20
= 1/5
====
2) Total number of outcomes = 20-1 ( 1 --> a not defective bulb was taken away in first trial and is not replaced. ) = 19
No of favourable outcomes = 19 - 4(defective) = 15
P(drawing a bulb which is not defective) = 15/19
======================================
HOPE IT HELPS!
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