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

Answer:

a) 11.6%

b) 0.52%

c) 1.75

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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

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