Question

Suppose that 60​% of the voters in a state intend to vote for a certain candidate. What is the probability that a survey polling 5 people reveals that at most 2 voters support the candidate.

Answer 1

Answer: 0.31744

Step-by-step explanation:

Binomial probability distribution formula :-

$P(X)=^nC_x \ p^x\ (1-p)^{n-x}$, where P(x) is the probability of getting success in x trials, n is total number of trials and p is the probability of getting succes in each trial.

Given : The probability that the voters in a state intend to vote for a certain candidate:  $p=0.60$

Now, the probability that a survey polling 5 people reveals that at most 2 voters support the candidate will be :-

$P(x\leq2)=P(0)+P(1)+P(2)\\\\=^5C_0 \ (0.60)^0\ (0.40)^{5}+^5C_1 \ (0.60)^1\ (0.40)^{4}+^5C_2 \ (0.60)^2\ (0.40)^{3}\\\\=(0.40)^5+5(0.60)(0.40)^4+10(0.60)^2(0.40)^3=0.31744$

Hence, the probability that a survey polling 5 people reveals that at most 2 voters support the candidate = 0.31744

Answer 2

Answer:

The probability that a survey polling 5 people reveals that at most 2 voters support the candidate is 0.08704.

Step-by-step explanation:

Given : Suppose that 60​% of the voters in a state intend to vote for a certain candidate.

To find : What is the probability that a survey polling 5 people reveals that at most 2 voters support the candidate.

Solution :

Applying binomial probability distribution,

$P(X=k)=\frac{n!}{k!(n-k)!}\times p^k\times (1-p)^{n-k}$

Here n=5 number of people

p=60%=0.6 probability of success.

k is at most 2 i.e. $k<2$

So, Probability is $P(X<2)=P(0)+P(1)$

The probability for k=0.

$P(X=0)=\frac{5!}{0!(5-0)!}\times 0.6^0\times (1-0.6)^{5-0}$

$P(X=0)=1\times 1\times 0.01024$

$P(X=0)=0.01024$

The probability for k=1.

$P(X=1)=\frac{5!}{1!(5-1)!}\times 0.6^1\times (1-0.6)^{5-1}$

$P(X=1)=\frac{5\times 4!}{4!}\times 0.6\times 0.0256$

$P(X=1)=5\times 0.6\times 0.0256$

$P(X=1)=0.0768$

Substitute in $P(X<2)=P(0)+P(1)$

$P(X<2)=0.01024+0.0768$

$P(X<2)=0.08704$

Therefore, The probability that a survey polling 5 people reveals that at most 2 voters support the candidate is 0.08704.