Question

students in a college math class counted 29 packages (1.5 ounces each) of plain m&ms and recorded the following color distribution.

color	number of m&ms
blue	300
green	165
yellow	231
brown	279
orange	269
total	1445

if one m&m is drawn at random from the total, find the probabilities.
a. the probability of getting a yellow candy.
b. the probability of getting a yellow or red candy.
c. the probability of not getting a yellow candy.
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Explanation:

Part a: Probability of yellow candy

Step 1: Identify counts

From the table, number of yellow M&Ms = 231, total M&Ms = 1445.

Step 2: Calculate probability

Probability = $\frac{\text{Number of yellow}}{\text{Total}} = \frac{231}{1445}$. Simplify (divide numerator and denominator by GCD, but 231 and 1445: 231 = 3×7×11, 1445 = 5×17×17. No common factors, so $\frac{231}{1445} \approx 0.1599$ (or keep as fraction).

Part b: Probability of yellow or red

Step 1: Identify counts

Yellow = 231, Red = 201, Total = 1445.

Step 2: Apply addition rule (mutually exclusive, so P(Yellow or Red) = P(Yellow) + P(Red))

P(Yellow) = $\frac{231}{1445}$, P(Red) = $\frac{201}{1445}$. Sum: $\frac{231 + 201}{1445} = \frac{432}{1445} \approx 0.2989$.

Part c: Probability of not yellow

Step 1: Use complement rule (P(not yellow) = 1 - P(yellow))

Step 2: Calculate

From part a, P(yellow) = $\frac{231}{1445}$, so 1 - $\frac{231}{1445}$ = $\frac{1445 - 231}{1445}$ = $\frac{1214}{1445} \approx 0.8401$.

Answer:

s:
a. $\frac{231}{1445} \approx 0.16$ (or exact fraction)
b. $\frac{432}{1445} \approx 0.30$ (or exact fraction)
c. $\frac{1214}{1445} \approx 0.84$ (or exact fraction)