Question

Carl has a piece of toast that has jelly on one side, and he dropped it twice. Both times, it landed with the jelly side up. If he drops it 2 more times, what is the probability that it will have landed jelly side up a total of 3 times? (A) 1/4(B) 1/3(C) 1/2(D) 2/3

Answer 1

Answer:

Option A - $\frac{1}{4}$

Step-by-step explanation:

Given : Carl has a piece of toast that has jelly on one side, and he dropped it twice. Both times, it landed with the jelly side up. If he drops it 2 more times.

To find : What is the probability that it will have landed jelly side up a total of 3 times?

Solution :  

Carl has a piece of toast which is two sided.

Probability of success is $p=\frac{1}{2}$

He drop the piece 4 times so n=4

The probability that it will have landed jelly side up a total of 3 times is given by binomial probability distribution.

i.e. $P(X=r)=^nC_r\times p^r\times (1-p)^{n-r}$

Here, r=3

$P(X=3)=^4C_3\times (\frac{1}{2})^3\times (1-\frac{1}{2})^{4-3}$

$P(X=3)=\frac{4!}{3!(4-3)!}\times (\frac{1}{2})^3\times (\frac{1}{2})^{1}$

$P(X=3)=4\times (\frac{1}{2})^4$

$P(X=3)=\frac{1}{4}$

Therefore, The probability that it will have landed jelly side up a total of 3 times is $\frac{1}{4}$

So, Option A is correct.