Question

As in Exercise \(\mathrm{P} .35,\) we have a bag of peanut \(\mathrm{M} \& \mathrm{M}\) 's with \(80 \mathrm{M} \& \mathrm{Ms}\) in it, and there are 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is yellow? (b) If we select one at random, what is the probability that it is not brown? (c) If we select one at random, what is the probability that it is blue or green? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are red? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is yellow and the second one is blue?

Short answer

(a) The probability that the M&M is yellow is 0.225, (b) The probability that it is not brown is 0.9, (c) The probability that it is blue or green is 0.3875, (d) The probability that both the first and second ones are red is 0.01895, and (e) The probability that the first one is yellow and the second one is blue is 0.057.

Step-by-step solution

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Calculate total number of M&M's in the bag. The total in the bag is the summation of the count of each color. That is \(80 = 11 (red) + 12 (orange) + 20 (blue) + 11 (green) + 18 (yellow) + 8 (brown)\). So there are 80 M&M's in total.

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(a) To find the probability that a randomly selected M&M is yellow, divide the number of yellow M&M's by the total number of M&M's. So, the probability is \(P(Yellow) = \frac{18}{80}\). It can be simplified as \(P(Yellow) = 0.225\).

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(b) To find the probability that a randomly selected M&M is not brown, subtract the proportion of brown M&M's from 1. So, the probability is \(P(Not Brown) = 1 - \frac{8}{80}\). Simplifying gives \(P(Not Brown) = 0.9\).

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(c) To calculate the probability that a randomly selected M&M is either blue or green, add the probabilities of selecting a blue and a green one. The probability is \(P(Blue or Green) = \frac{20}{80} + \frac{11}{80}\). Simplifying gives \(P(Blue or Green) = 0.3875\).

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(d) Since the picking is performed with replacement, the two events are independent. Therefore, the probability of both the first and second ones are red is simply the product of the probabilities of each individual event. So, \(P(Red, Red) = \left(\frac{11}{80}\right)^2\). Simplifying gives \(P(Red, Red) = 0.01895\).

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(e) In this case the selections are dependent. The probability that the first one is yellow and the second one is blue is the product of the probability of first yellow and the probability of blue given that one yellow has been taken out. So, \(P(Yellow, Blue) = \frac{18}{80} \cdot \frac{20}{79}\). Simplifying gives \(P(Yellow, Blue) = 0.057\).

Key concepts

Random Selection

In the realm of probability, random selection refers to the process where each item in a population has an equal chance of being chosen. Imagine a bag filled with M&M's of different colors, as described in the exercise. If you reach into the bag without looking, each M&M has the same probability of being picked, whether it's yellow, red, or blue. This scenario is a perfect example of a random selection.

For instance, to find the probability of pulling a yellow M&M from the bag, you would count how many yellow M&M's there are and divide that by the total number of M&M's. This straightforward method works because of the equal likelihood of selecting each candy. Providing an interactive simulation or visual representation can further illustrate this concept to students, enhancing their understanding by letting them virtually 'pull' M&Ms out of the bag multiple times to see the random selection principle in action.

Independent Events

When we talk about independent events in probability, we're referring to scenarios where the outcome of one event does not affect the outcome of another. In the M&M's exercise, selecting an M&M and then putting it back in the bag before selecting another one is an example of independent events. No matter what color you pick first, it won't impact the color you might pick next because the selection starts afresh with the full assortment of M&M's.

To calculate the probability of two independent events both occurring, like picking two red M&M's in a row with replacement, you would multiply the probability of the first event by the probability of the second event. This is why the probability of getting two red M&M's in two independent selections is \(P(Red, Red) = \left(\frac{11}{80}\right)^2\). Using real-life examples, like flipping a fair coin multiple times, can help students internalize the concept of independent events—no matter how many times you flip a coin, the odds of getting heads or tails are always the same.

Dependent Events

Conversely, dependent events occur when the outcome of one event affects the outcome of another. If you pick an M&M from the bag and do not replace it before picking another one, the outcomes are dependent. Removing an M&M alters the total number and the composition of the candies remaining in the bag, thereby changing the probabilities of the subsequent selection.

In the exercise, after one yellow M&M is taken out, there is one less candy to choose from when selecting the second one, which is now more likely to be a different color than the first. The probability that the first candy is yellow and the second one is blue is calculated by multiplying the chance of picking a yellow (with all M&M's present) by the probability of then picking a blue with the yellow one removed: \(P(Yellow, Blue) = \frac{18}{80} \cdot \frac{20}{79}\). Using guided exercises where students can remove items from a set and see the influence on subsequent choices can effectively demonstrate this concept of dependency in probability.