Question

About 60% of the car accidents in a given highway road are due to exceed of the limited high speed. If a study will investigate 10 cars accidents

1. The probability that none is due to exceed of the limited high speed is
2. The probability that 5 exact accidents are due to exceed of the limited high speed is
3. The probability that at least 1 car accident is due to exceed of the limited high speed is
4. The probability that at most 1 car accident is due to exceed of the limited high speed is
5. The expected number of car accidents due to exceed of the limited high speed is

Answer 1

We can consider this experiment as BINOMIAL EXPERIMENT because of the following :

1) n= 10 car accidents (i.e number of trials) is fixed.

2) Accidents ( trials ) are independent of one another and they are identical.

3) Considering each accident (trial) it is because of either exceed of limited high speed (success) or some other cause(failure)

4) Probably of a car accident due to exceed of limited high speed i.e probability of success =60% = 0.6 is same for every accident.

Now we have to find the following :

1) let. X be the number of accidents due to exceed of limited high speed.

we need to get P(X=0)

FORMULA: P(X=x) = ncx . p^x . q^n_x

here n= 10, x= 0, p= 0.6, q=1_0.6 =0.4

p(X=0) = 10c0 . (0.6)^0 . (0.4)^10 =0.0001 answer

Probability that none is due to exceed of limited high speed = 0.0001

2) P(X=5) = 10c5 . (0.6)^5 . (0.4)^10_5 = 0.2007 answer

Probability that exactly 5 accidents are due to exceed of limited high speed = 0.2007

3) P(X 1) = P(X=1)+P(X=2)+.........+P(X=10)

=1_P(X=0) = 1_0.0001 = 0.999

probability that at least one car accident is

due to exceed of high speed limit = 0.9999

4)P(X1) = P(X=0)+P(X=1) =0.0001 + 10c1 . (0.6)^1 . (0.4)^9 = 0.0017

Probability that Atmost one car accident is due to the exceed of limited high speed = 0.0017

5) Expected number of car accidents due to exceed of limited high speed. = Mean

= n. p= 10. (0.6)

= 6 answer