Question

Joe is new to cooking. The probability that joe burns dinner is 50%

Answer 1

Given that the probability that Joe burns dinner is:

`p=50\text{\%}=(50)/(100)=0.5`

You need to use the following Binomial Distribution Formula, in order to solve the exercise:

`P(X)=(n!)/((n-x)!x!)\cdot p^x(1-p)^(n-x)`

Where "n" is the sample size, "x" is the number of successes desired, and "p" is the probability of getting a success in a trial.

(a) You must find the probability that in the next 7 dinners Joe prepares, 4 of them will burn. Therefore, for this case:

`\begin{gathered} n=7 \\ x=4 \end{gathered}`

Then, you can substitute values into the formula and evaluate:

`P(X=4)=(7!)/((7-4)!4!)(0.5)^4(1-0.5)^(7-4)`
`P(X=4)\approx0.2734`

(b) You must find the probability that in the next 7 dinners Joe prepares, at least 5 of them will burn. Therefore, since "at least" indicates greater than or equal to 5, you need to set up this Sum for:

`\begin{gathered} x=5 \\ x=6 \\ x=7 \end{gathered}`

Then:

`P(X\ge5)=(7!)/((7-5)!5!)(0.5)^5(1-0.5)^(7-5)+(7!)/((7-6)!6!)(0.5)^6(1-0.5)^(7-6)+(7!)/((7-7)!7!)(0.5)^7(1-0.5)^(7-7)`

Evaluating, you get:

`P(X\ge5)\approx0.2266`

(c) You must find the probability that in the next 7 dinners Joe prepares, less than 3 of them will burn. Therefore, you need to set up a Sum for:

`\begin{gathered} x=0 \\ x=1 \\ x=2 \end{gathered}`

As follows:

`P(X<3)=(7!)/((7-0)!0!)(0.5)^0(1-0.5)^(7-0)+(7!)/((7-1)!1!)(0.5)^1(1-0.5)^(7-1)+(7!)/((7-2)!2!)(0.5)^2(1-0.5)^(7-2)`

Evaluating, you get:

`P(X<3)\approx0.2266`

Hence, the answers are:

(a)

`0.2734`

(b)

`0.2266`

(c)

`0.2266`