Ten eggs are drawn successively with replacement from a lot containing 10 % defective eggs. Then, the probability that there is atleast one defective egg.

Question

Ten eggs are drawn successively with replacement from a lot containing 10 % defective eggs. Then, the probability that there is atleast one defective egg. (A) 1-(710/1010) (B) 1+(710/1010) (C) 1+(910/1010) (D) 1-(910/1010)

Tardigrade Admin · Accepted Answer

Let X denote the number of defective eggs in the 10 eggs drawn. Since, the drawing is done with replacement, the trials are Bernoulli trials.Clearly, X has the binomial distribution with n=10 and p=(10/100)=(1/10) Therefore, q=1-p=(9/10) Now, P (atleast one defective egg) =P(X ≥ 1)=1-P(X=0) =1- 10 C0((9/10))10=1-(910/1010)

Q. Ten eggs are drawn successively with replacement from a lot containing $10%$ defective eggs. Then, the probability that there is atleast one defective egg.

$1 - \frac{7 ^{10}}{1 0 ^{10}}$

$1 + \frac{7 ^{10}}{1 0 ^{10}}$

$1 + \frac{9 ^{10}}{1 0 ^{10}}$

$1 - \frac{9 ^{10}}{1 0 ^{10}}$

Q. Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Then, the probability that there is atleast one defective egg.

1−1010710​

1+1010710​

1+1010910​

1−1010910​

Q. Ten eggs are drawn successively with replacement from a lot containing $10%$ defective eggs. Then, the probability that there is atleast one defective egg.

$1 - \frac{7 ^{10}}{1 0 ^{10}}$

$1 + \frac{7 ^{10}}{1 0 ^{10}}$

$1 + \frac{9 ^{10}}{1 0 ^{10}}$

$1 - \frac{9 ^{10}}{1 0 ^{10}}$