Write the probability distribution of the defective bulbs drawn in three draw when it is known that in a lot of 10 on an average 3 are defective
Anonymous:
RS Agarwal??
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step-by-step explanation:
let X denotes the random variable showing the number of defective bulbs.
then,
X can take the values 0,1,2 or 3.
therefore,
P(X=0) = P( none bulbs are defective)
= P(all the 3 bulbs are good)
= c(7,3)/c(10,3)
= (7×6×5×3×2×1)/(3×2×10×9×8)
= 7/24
now,
P(X=1) = P(1 defective and 2 are good ones)
= c(3,1) × c(7,2)/c(10,3)
= (3×7×6×3×2)/(2×10×9×8)
= 21/40
now,
P(X=2) = P(2 are defective and 1 good one)
= c(3,2) × c(7,1)/c(10,3)
= (3×2×7×3×2)/(2×10×9×8)
= 7/40
also,
P(X=3) =P(3 defective bulbs)
= c(3,3)/c(10,3)
= (3×2)/(10×9×8)
= 1/120
Thus,
the probability distribution ia given by:-
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.
.
.
.
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Note:-
Kindly See the attachment
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Answer:
your answer will be 3.333333 because.
total no of bulbs be 10
on an average defective bulbs are 3
probability formula:sum of observation
____________
total no of observations
3/10
=3.333333
#50%sure ANSWER
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