Question

1. you randomly choose one cube from group 1 and one pyramid from group 2. what is the probability that you will choose either an unshaded cube or an unshaded pyramid?

group 1
group 2
o $\frac{62}{77}$
o $\frac{68}{77}$
o $\frac{53}{77}$
o $\frac{15}{77}$
clear all

Explanation:

Step1: Count the number of items in each group

Group 1 has 14 cubes, 6 of which are un - shaded. Group 2 has 11 pyramids, 5 of which are un - shaded.

Step2: Calculate the probability of choosing an un - shaded cube

The probability of choosing an un - shaded cube $P(C_{unshaded})=\frac{6}{14}=\frac{3}{7}$.

Step3: Calculate the probability of choosing an un - shaded pyramid

The probability of choosing an un - shaded pyramid $P(P_{unshaded})=\frac{5}{11}$.

Step4: Calculate the probability of choosing both an un - shaded cube and an un - shaded pyramid

Since the events of choosing a cube and a pyramid are independent, $P(C_{unshaded}\cap P_{unshaded})=\frac{3}{7}\times\frac{5}{11}=\frac{15}{77}$.

Step5: Use the addition rule of probability

The addition rule for two events $A$ and $B$ is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here $A$ is the event of choosing an un - shaded cube and $B$ is the event of choosing an un - shaded pyramid. So $P(C_{unshaded}\cup P_{unshaded})=\frac{3}{7}+\frac{5}{11}-\frac{15}{77}$. First, find a common denominator of 77. $\frac{3}{7}=\frac{3\times11}{7\times11}=\frac{33}{77}$ and $\frac{5}{11}=\frac{5\times7}{11\times7}=\frac{35}{77}$. Then $\frac{33}{77}+\frac{35}{77}-\frac{15}{77}=\frac{33 + 35-15}{77}=\frac{53}{77}$.

Answer:

$\frac{53}{77}$