Home › MHT-CET › Mathematics › Binomial Distribution › Ten eggs drawn with replacement from a lot of 10…
Ten eggs drawn with replacement from a lot of 10% defective. Probability that **at least one** is defective:
A$1 - (9/10)^{10}$
B$(1/10)^{10}$
C$10 \cdot (1/10)$
D$10!$
Answer & Solution
Correct answer: A. $1 - (9/10)^{10}$
$P(X \geq 1) = 1 - P(X = 0) = 1 - {}^{10}C_0 (1/10)^0 (9/10)^{10} = 1 - (9/10)^{10} \approx 1 - 0.349 = 0.651$.
Related questions
A multiple-choice exam has 10 questions each with 5 options. The probability that a studenA bulb-factory has 5% defectives. In a random sample of 10 bulbs, the probability of **notIn a 5-trial binomial distribution, $P(X = 1) = 0.4096$ and $P(X = 2) = 0.2048$. The probaIf success probability in a single trial is 0.01, the minimum number of trials needed so tProbability of bomb hitting target = 0.8. Out of 10 bombs dropped, probability that **exacIf $X im B(4, p)$ and $P(X = 0) = 16/81$, then $P(X = 4)$ equals:Mean and variance of a binomial distribution are 18 and 12 respectively. The value of $n$ The probability that a shooter hits a target is 3/4. The minimum number of shots required