Three statistics professors and seven chemistry professors are available to be advisors to a student organization. The student organization needs two of the professors to be advisors. If each professor has an equal chance of being selected, what is the probability that both professors are chemistry professors?

P[2qp] = 0,467      or    46,7 %

Step-by-step explanation:

The probability of two of seven chemistry professors is equal to:

P[2 qp] = quantity of combinations of seven elements taken in groups of two, over total quantity of events, therefore

The total quantity of combinations is:

3 statistics professors + 7 chemistry professors taken in groups of 2 that is:

Te  = (3 + 7) ! / 2!*(10-2)!

Te = 9!/ 2!*5!        ⇒    Te = 9*8*7*6*5!/2!*8!    ⇒ Te = 9*8!/2*8!

Te = 45

And successful events are:

C₇,₂  =  7!/2!*(7-2)!

C₇,₂  = 21

Finally, the probability that both professors are chemistry professors is:

P[2qp] = 21/45

P[2qp] = 0,467      or    46,7 %

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol