A pizzeria offers 8 different ingredients on its pizza. Andy chooses 3 different ingredients from the 8 by drawing the names from a hat. What is the probability that Andy chooses either mushrooms or pepperoni or both among his three ingredients?
The answer is 9/14, but how do you get it?
You can do it this way:
Pr(either m or P or both)=1-Pr(neither M or P)
Pr(neither)=6/8*5/7*4/6= 120/336=40/112=20/56=10/28= 5/14
Pr(either m or P or both)=1-5/14=9/14
Thank you! I always forget you can minus the complement from 1.
To solve this problem, we can use the principle of inclusion-exclusion.
First, we need to find the probability of Andy choosing mushrooms or pepperoni or both.
Step 1: Find the probability of Andy choosing mushrooms.
Out of the 8 ingredients, one of them is mushrooms. Since Andy is choosing 3 ingredients, we can choose the mushrooms in 3 ways:
(1) He chooses only mushrooms.
(2) He chooses mushrooms along with 2 other ingredients.
(3) He chooses mushrooms along with 1 other ingredient.
The probability of choosing only mushrooms is ${1 \choose 1}$ / ${8 \choose 3} = 1 / 56$, as there is only 1 way to choose mushrooms and 3 ingredients in total.
The probability of choosing mushrooms along with 2 other ingredients is ${1 \choose 1}$ / ${8 \choose 3}$ = 1 / 56, as there is only 1 way to choose mushrooms and 3 ingredients from 8.
The probability of choosing mushrooms along with 1 other ingredient is ${1 \choose 1}$ / ${8 \choose 3}$ = 1 / 56, as there is again only 1 way to choose mushrooms and 3 ingredients from 8.
Step 2: Find the probability of Andy choosing pepperoni.
Similarly, out of the 8 ingredients, one of them is pepperoni. We can choose the pepperoni in 3 ways:
(1) He chooses only pepperoni.
(2) He chooses pepperoni along with 2 other ingredients.
(3) He chooses pepperoni along with 1 other ingredient.
The probability of choosing only pepperoni is ${1 \choose 1}$ / ${8 \choose 3} = 1 / 56$.
The probability of choosing pepperoni along with 2 other ingredients is also ${1 \choose 1}$ / ${8 \choose 3}$ = 1 / 56.
The probability of choosing pepperoni along with 1 other ingredient is ${1 \choose 1}$ / ${8 \choose 3}$ = 1 / 56.
Step 3: Find the probability of Andy choosing both mushrooms and pepperoni.
Andy can choose both mushrooms and pepperoni along with 1 other ingredient. In this case, there is only 1 way to choose both mushrooms and pepperoni, and again ${1 \choose 1}$ / ${8 \choose 3}$ = 1 / 56.
Step 4: Calculate the total probability.
To find the total probability, we need to add the probabilities from Step 1, Step 2, and subtract the probability from Step 3 to avoid double counting.
Total probability = (probability of choosing mushrooms) + (probability of choosing pepperoni) - (probability of choosing both mushrooms and pepperoni)
Total probability = [1 / 56 + 1 / 56 + 1 / 56] + [1 / 56 + 1 / 56 + 1 / 56] - (1 / 56)
Total probability = 6 / 56 - 1 / 56
Total probability = 5 / 56
Simplifying the fraction, we get 1 / 11.
Therefore, the probability that Andy chooses either mushrooms or pepperoni or both among his three ingredients is 1 / 11.
Note: The answer you mentioned (9/14) seems incorrect based on the given problem.