Question

The probability that a mature hen will lay an egg on a given day is 0.80. Hannah has 12 hens. What is the probability that 10 of the 12 hens will lay eggs on a given day? 0.05 0.11 0.20 0.28

Answer 1

Answer: 0.28

Step-by-step explanation:

Binomial probability formula :

$P(X=x)=\ ^nC_xp^x(1-p)^x$ , where  x = Number of success , n= sample size , p= probability of success for each trial.

Let x = Number of hens lay eggs.

then, p=  0.80

n= 12

For x=10

$P(x=10)=\ ^{12}C_{10}(0.80)^{10}(0.20)^{2}\\\\=\dfrac{12!}{10!2!}\times 0.1073741824\times 0.04\\\\\approx0.28$

Hence, the required probability = 0.28

Answer 2

The probability that 10 of the 12 hens will lay eggs on a given day is 0.28.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consist of n independent Bernoulli trials.

Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

Binomial probability formula :

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

where x = Number of successes,

n= sample size,

p= probability of success for each trial.

Let x be the number of hens that lay eggs.

then, p =  0.80

n= 12

For x=10

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}\\\\P(X =x) = \: ^{12}C_{10}p^{10}(1-p)^{{12}-{10}}\\\\P(X =x) = 12! /10!2! \times 0.10737 \times 0.04\\\\P(X =x) = 0.28$

Hence, the required probability is 0.28.

Learn more about binomial distribution here:

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