Question

1.4b hw: probability day 2
find the probability.

1. a single card is chosen at random from a standard deck of 52 playing cards. what is the probability of choosing a card that is not a diamond?
2. if a single 6 - sided die is rolled, what is the probability of rolling a number that is not 1 or 5?
3. in a college science class there are eight students. three are juniors and five are seniors. if three students are selected, find the probability that at least one is a senior.
4. the triangle shown to the right is made up of all equilateral triangles. what is the probability that a randomly chosen point in the triangle is not in the shaded region?

Explanation:

Step1: Calculate number of non - diamond cards

A standard deck has 52 cards, and 13 are diamonds. So the number of non - diamond cards is $52 - 13=39$.
The probability formula is $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Here, the total number of outcomes is 52 and the number of favorable outcomes (non - diamond cards) is 39. So the probability of choosing a non - diamond card is $\frac{39}{52}=\frac{3}{4}$.

Step2: Calculate probability for the die

A 6 - sided die has 6 possible outcomes. The numbers that are 1 or 5 are 2 outcomes. So the numbers that are not 1 or 5 are $6-(2)=4$.
Using the probability formula $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$, the probability is $\frac{4}{6}=\frac{2}{3}$.

Step3: Calculate probability for students

First, find the probability of the complement event (no seniors are selected).
The total number of ways to select 3 students out of 8 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 8$ and $r = 3$. So $C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8\times7\times6}{3\times2\times1}=56$.
The number of ways to select 3 students from 3 juniors (i.e., no seniors) is $C(3,3)=\frac{3!}{3!(3 - 3)!}=1$.
The probability of selecting no seniors is $\frac{C(3,3)}{C(8,3)}=\frac{1}{56}$.
The probability of at least one senior is $1-\frac{1}{56}=\frac{55}{56}$.

Step4: Calculate probability for the triangle

Assume the large triangle is made up of 9 equal - sized equilateral sub - triangles and 1 is shaded.
The probability of a point being in the shaded region is $\frac{1}{9}$.
The probability of a point not being in the shaded region is $1-\frac{1}{9}=\frac{8}{9}$.

Answer:

1. $\frac{3}{4}$
2. $\frac{2}{3}$
3. $\frac{55}{56}$
4. $\frac{8}{9}$