Question

the numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement. find the given probabilities.
a. the sum of the numbers is 8.
b. the sum of the number is 4 or less
c. the first number is 2 or the sum is 5.
a. the probability that the sum of the numbers is 8 is $\frac{1}{10}$ (type an integer or a simplified fraction.)
b. the probability that the sum of the numbers is 4 or less is $\frac{1}{5}$ (type an integer or a simplified fraction.)
c. the probability that the first number is 2 or the sum is 5 is (type an integer or a simplified fraction.)

Explanation:

Step1: Calculate total number of ways to draw 2 - slips

The number of ways to draw 2 slips out of 5 without replacement is given by the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 5$ and $r=2$. So, $P(5,2)=\frac{5!}{(5 - 2)!}=\frac{5!}{3!}=5\times4 = 20$ ways.

Step2: Define events for part c

Let $A$ be the event that the first - number is 2. The number of outcomes where the first number is 2 is 4 (i.e., (2,1), (2,3), (2,4), (2,5)). Let $B$ be the event that the sum of the two numbers is 5. The pairs that sum to 5 are (1,4), (4,1), (2,3), (3,2), so there are 4 outcomes. The intersection of $A$ and $B$ is the set of outcomes that are in both $A$ and $B$, which is the pair (2,3), so $n(A\cap B)=1$.

Step3: Use the addition rule of probability

The addition rule of probability is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Since $P(A)=\frac{4}{20}$, $P(B)=\frac{4}{20}$, and $P(A\cap B)=\frac{1}{20}$, then $P(A\cup B)=\frac{4 + 4-1}{20}=\frac{7}{20}$.

Answer:

$\frac{7}{20}$