Question

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

Answer 1

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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:-

.

.

.

.

.

.

Note:-

Kindly See the attachment

Answer 2

Answer:

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