Cranky mower To start her old lawn mower, Rita has to pull a cord and hope for some luck. On any particular pull, the mower has a 20% chance of starting. (a) Find the probability that it takes her exactly 3 pulls to start the mower. Show your work. (b) Find the probability that it takes her more than 10 pulls to start the mower. Show your work.

Respuesta :

Answer:

(a) = 0.128

(b) = 0.1074

Step-by-step explanation:

Probability of starting the mower = 20%

Probability of  not starting the mower = 80%

(a) In order to take her exactly three tries, the mower should not turn on with the first two attempts but turn on with the third attempt:

[tex]P(n=3) = 0.80*0.80*0.20=0.128[/tex]

(b) The probability that it takes her more than 10 pulls to start the mower is given by the probability of 10 straight failed attempts happening:

[tex]P(n>10) = 0.80^{10}=0.1074[/tex]

Using the binomial distribution, it is found that there is a

a) 0.128 = 12.8% probability that it takes her exactly 3 pulls to start the mower.

b) 0.1074 = 10.74% probability that it takes her more than 10 pulls to start the mower.

For each pull, there are only two possible outcomes. Either the mower starts, or it does not. The probability of the mower starting on a pull is independent of any other pull, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

On each particular pull, it has a 20% chance of starting, hence [tex]p = 0.2[/tex]

Item a:

This probability is it doesn't start on the first two, which is P(X = 0) when n = 2, multiplied by 0.2, which is the probability it starts on the third. Hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{2,0}.(0.2)^{0}.(0.8)^{2} = 0.64[/tex]

[tex]p = 0.64(0.2) = 0.128[/tex]

0.128 = 12.8% probability that it takes her exactly 3 pulls to start the mower.

Item b:

This probability is none on the first 10, which is P(X = 0) when n = 10, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074[/tex]

0.1074 = 10.74% probability that it takes her more than 10 pulls to start the mower.

A similar problem is given at https://brainly.com/question/24863377