Question

In a state where license plates contain six digits, what is the probability that the license number of a randomly selected car has exactly two 9's? Assume that each digit of the license number is randomly selected from .

Answer 1

Answer:

Step-by-step explanation:

Given : In a state where license plates contain six digits.

Probability of that a number is 9 = $\dfrac{1}{10}$   [Since total digits = 10]

We assume that each digit of the license number is randomly selected  .

Since each digit in the license plate is independent from the other  and there is only two possible outcomes for given case (either 9 or not), so we can use Binomial.

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

, where n= total trials , p = probability for each success.

Let x be the number of 9s in the license plate number.

$X\sim Bin(n=6, p=\dfrac{1}{10}})$

Then, the probability that the license number of a randomly selected car has exactly two 9's will be :

$P(X=2)=^6C_2(\dfrac{1}{10})^2(1-\dfrac{1}{10})^{6-2}\\\\=\dfrac{6!}{2!(6-2)!}(\dfrac{1}{100})(\dfrac{9}{10})^4=0.098415$

Hence, the required probability = 0.098415