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:

$p=50\text{\%}=\frac{50}{100}=0.5$

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

$P(X)=\frac{n!}{(n-x)!x!}\cdot p^x(1-p)^{n-x}$

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)=\frac{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:

$P(X\ge5)=\frac{7!}{(7-5)!5!}(0.5)^5(1-0.5)^{7-5}+\frac{7!}{(7-6)!6!}(0.5)^6(1-0.5)^{7-6}+\frac{7!}{(7-7)!7!}(0.5)^7(1-0.5)^{7-7}$

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:

$P(X<3)=\frac{7!}{(7-0)!0!}(0.5)^0(1-0.5)^{7-0}+\frac{7!}{(7-1)!1!}(0.5)^1(1-0.5)^{7-1}+\frac{7!}{(7-2)!2!}(0.5)^2(1-0.5)^{7-2}$

Evaluating, you get:

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

Hence, the answers are:

(a)

$0.2734$

(b)

$0.2266$

(c)

$0.2266$