This past semester, a professor had a small business calculus section. The students in the class were Jinita comma Al comma Jim comma and William. Suppose the professor randomly selects two people to go to the board to work problems. What is the probability that Jim is the first person chosen to go to the board and Al is the second?

Permutation helps us calculate the number of ways to order items. Basically we multiply the number of elements that can be placed first, then for second place until there is no empty place.

In this case, the probability of selecting Jim first is 1 over the total number of students that is 4. Then:

[tex]\frac{1}{4}[/tex]

To select Al in second place is similar, but this time we have 3 students since Jim has already been placed.

[tex]\frac{1}{3}[/tex]

Thus, in total the probability will be:

[tex](\frac{1}{4} )(\frac{1}{3} )=\frac{1}{12} =0.0833[/tex]

jtellezd jtellezd · Answer 2 · 2019-10-17T21:43:04+02:00

Answer:

P = 1/12

Step-by-step explanation:

The probablity that two dependent events A and B occur is

P ( A y B ) = P(A) * P ( B | A)

According to P( A) = favorables events/ total possible events

P (A) (probability of choosing Jim) favorable events 1

Total possible events 4

P (A) = 1/4

Then the probability of Al is: P( B )

P ( B ) (probability of choosing Al) favorable events 1

Total possible events 3

P ( B ) = 1/3

Then the probability of chossing Al having selected Jim first is 1/12

P( B ) =