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

5 0

3 years ago

If the original quantity is 10 and the new quantity is 14, what is the percent increase?

Korvikt [17]

Answer:

40% increase

Step-by-step explanation:

10 is you 100%

if it increased to 14, then that means your 100% = 10 increased +4 which is the 40%

6 0

3 years ago