Question

(a) Would you be surprised if none of the candies were orange? Compute an appropriate probability to support your answer.

(b) How surprising would it be to get $5$or more orange candies? Compute an appropriate probability to support your answer

Short answer

(a)The probability is$0.1678$

(b) The probability is$0.0104$

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that, according to the Mars candy company, $20\%$of its plain M&M’s candies are orange. Assume that the company’s claim is true. Suppose that you reach into a large bag of plain M&M’s (without looking) and pull out $8$candies. Let$X$ = the number of orange candies you get.

Part (a) Step 2: Explanation

Given:

Probability of success $\left(p\right)=20\%=0.20$.

Number of trials $\left(n\right)=8$

Calculation:

Here, $\mathrm{X}$is a random variable that indicates the number of orange candies.

The probability of selecting $0$orange candies is:

$P(X=0){=}^8{C}_0\times (0.20{)}^0\times (1-0.20{)}^{8-0}$

$=0.1678$

Thus, the probability is$0.1678.$

Here, probability is above$5\%$. Thus, the event is not surprising.

Part (b) Step 1: Given information

Given in the question that, According to the Mars candy company, $20\%$ of its plain M&M’s candies are orange. Assume that the company’s claim is true. Suppose that you reach into a large bag of plain M&M’s (without looking) and pull out 8 candies. Let $X$ = the number of orange candies you get.

Part (b) Step 2: Explanation

The probability of selecting more than $5$orange candies is calculated as:

$P(X\ge 5)=1-P(X<5)$

$=1-\left[P\right(X=0)+P(X=1)+\dots .+P(X=4\left)\right]$

$=0.0104$

Thus, the probability is $0.0104$.

Here, probability is below $5\%$. Thus, the event is surprising to happen.