The question is as follows: Your company is seeking to tender for work with three clients: A, B & C. It estimates that the probability that it will secure tender A is 0.2, tender B is 0.3 and tender C is 0.4. Find the probability that your company will:
(a) win all 3 tenders
(b) win only one tender
Help!!!!!!!
2007-07-16 07:01:40 · 4 answers · asked by brief reunion 1 in Science & Mathematics ➔ Mathematics
please show working out.
2007-07-16 07:02:26 · update #1
P(all 3) = 0.2*0.3*0.4 = 0.024
P(only 1) = 0.2*0.7*0.6 + 0.8*0.3*0.6 + 0.8*0.7*0.4
= 0.452
2007-07-16 07:13:24 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
A. This one is easy. You just have to multiply the chances together. In this case the result is 0.024, or 2.4% (1 in 41.666 chance).
B. This one is a little harder. What you need to do is find the probability for each tender that it will be one and that the other ones will not be won, and then add those chances together. In this case we have:
0.2*(1-0.3)*(1-0.4)=0.084
0.3*(1-0.2)*(1-0.4)=0.144
0.4*(1-0.2)*(1-0.3)=0.224
0.084+0.144+0.224=0.452, or 45.2% (1 in 2.2124 chance).
2007-07-16 07:08:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
a. 0.2*0.3*0.4 = 0.024 or 2.4%
I'm stumped on b).
2007-07-16 07:09:05 · answer #3 · answered by gebobs 6 · 0⤊ 0⤋
all 3, P(A)*P(B)*P(C)
(.2)(.3)(.4)= .024
2007-07-16 07:07:17 · answer #4 · answered by clevelandbrownsgirl2007 3 · 1⤊ 0⤋