Question

Twonumbersarerandomlyselectedwithoutreplacementfrom{1,2,3,4,5}.Findtheprobabilitythat:a) Bothareeven.b) Bothareprime.c) Thesumofthe numbersis odd.d) Theproductofthenumbersis odd.

Answer 1

Given the numbers:

{1,2,3,4,5}

If two numbers are randomly selected without replacement, let's find the following:

a) Both are even

Number of even numbers in {1,2,3,4,5} = 2 and 4 = two even numbers.

Probability that even numbers are picked without replacement is:

$P(\text{even)}=\frac{2}{5}\times\frac{1}{4}=\frac{2}{20}\text{ = }\frac{1}{10}$$P(\text{even)}=\frac{1}{10}$

b) Both are prime:

Number of prime numbers in {1,2,3,4,5} = 1, 3 and 5 = three prime numbers

Probability that the random umbers picked are prime numbers is:

$\begin{gathered} P(both\text{ prime)=}\frac{3}{5}\times\frac{2}{4}=\frac{6}{20}=\frac{3}{10} \\ \\ P(\text{both prime)=}\frac{3}{10} \end{gathered}$

c) The sum of the numbers is odd.

we have the following:

1 + 2

1 + 4

2 + 3

2 + 5

3 + 4

4 + 5

Here, we have 6 possible sets where the sum of the numbers is odd.

Total number of sets using =

$^5C_2=\frac{5!}{2!(5-2)!}=\frac{5!}{2!(3)!}=10$

Thus, we have:

$\begin{gathered} P\text{ = }\frac{6}{10}=\frac{3}{5} \\ \\ P\text{ = }\frac{3}{5} \end{gathered}$

d) The product of the numbers is odd:

1 x 3

1 x 5

3 x 5

$P=\frac{3}{10}$

ANSWER:

$a)\text{ P(both are even) = }\frac{1}{10}$$b)\text{ P(both are prime) = }\frac{3}{10}$$c)\text{ P(odd sum) = }\frac{3}{5}$$d)\text{ P(odd product)=}\frac{3}{10}$