Question

There are 5 traffic signals between your home and work. Each is red with probability 0.35 , independently of all others. Find: a) the probability of encountering no red lights, 2.26% 5.2% 11.6% 17.5% unanswered b) the probability of encountaring only red lights, 0.03% 0.52% 1.16% 16.4% unanswered c) the expected number of red lights you will encounter? 0.75 1.42 1.75 2.25 unanswered SubmitYou have used 0 of 4 attemptsSome problems have options such as save, reset, hints, or show answer. These options follow the Submit button.

Answer 1

a) 11.6%

b) 0.52%

c) 1.75

Explanation:

For each light, there are only two possible outcomes. Either it is red, or it is not. The probability of a light being red is independent from other lights. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

There are 5 traffic signals between your home and work.

This means that
`n = 5`

Each is red with probability 0.35

This means that
`p = 0.35`

a) the probability of encountering no red lights

This is P(X = 0). So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(5,0).(0.35)^(0).(0.65)^(5) = 0.116`

So the answer for a is 11.6%

b) the probability of encountaring only red lights

This is P(X = 5)

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 5) = C_(5,5).(0.35)^(5).(0.65)^(0) = 0.0052`

So the answer for b is 0.52%.

c) the expected number of red lights you will encounter?

The expeced number of the binomial distribution is given by:

`E(X) = np`

So

`E(X) = 5*0.35 = 1.75`