Question

If 3 of 12 car drivers do not carry driving license, what is the probability that a traffic inspector who

randomly checks 3 car drivers, will catch (i) 1 for not carrying driving license (ii) at least 2 for not

carrying driving license?
​

Answer 1

Answer:

Step-by-step explanation:12 car drivers so n=12

success probability is p=3/12

so q=3/4

(i) p(x=1)=12^c(1/4)^1 (3/4)^11

              =0.1267

Answer 2

Answer:

the required probability is $\frac{1}{4}$  and   $\frac{1}{3}$ respectively.

Step-by-step explanation:

here as per the given question, we have given that out of 12 , 3 drivers do not carry license.

so we have to find out that if a traffic inspector checks for license for 3 car drivers , what is the probability that  one of them get caught for not carrying the license.

so we know that probability is the ratio of all the favourable outcomes to the total possible outcomes.

so in this case ,

favorable outcome is $^3C_1$= 3

and there are total 12 drivers, so the probability that 1 will be caught in 3 random,

probability (i)=

$\frac{3}{12}=\frac{1}{4}$

and for the second case, there are atleast 2 people for not carrying the license.

so there can be either 2 or 3 people caught, so the required probabilty is

$\frac{^3C_2}{12} +\frac{^3C_3}{12}\\\\ \frac{3+1}{12} \\\frac{1}{3}$

P(ii)=$\frac{1}{3}$

for more such questions, you can visit:https://brainly.in/question/16829693

#SPJ2