Probability help

[archived_user]

Can anyone please explain why they didnt do just (0.25)(0.42) below?As far as I know joint probability of independent events is dictated by the formula P(AB) = P(A) X P(B).Why did they subtract the cumulative probability of all 3 events from the joint probability of A and C ?

 

[archived_user]

You’re correct, as far as you go. Here, you have three events that you want to occur together: A, ~B, and C. So the solution is:
P(A & ~B & C) = P(A)P(~B)P(C) = (0.25)(1 – 0.15)(0.42) = 0.0893 = 8.93%

 

[archived_user]

(1) Take the probability that A & B happen together: .25 *.15 = .105
(2) Take the probability that A & B & C all happen together: .25 * .15. * .42 = .01575
(3) Subtract (2) from (1): .105 - 01575 = .08925

 

[archived_user]

Thanks S2000 and JSD.That cleared it up

 

[archived_user]

My pleasure.