Thread: Probability of two events

A company believes there is a 40% probability of winning the first contract. If they win the first contract, the probability of winning the second is 70%. However, if they lose the first contract, the president thinks that the probability of winning the second contract decreases to 50%.

a) What is the probability that they win both contracts?
b) What is the probability that they lose both?
c) what is the probability that they win only one contract?

a) do I add the two chances 70+40 then divide it by 200?
b) same as A but with the losing percentages?
c)

Can someone help me with these ? Thank you.

Originally Posted by xfyz

A company believes there is a 40% probability of winning the first contract. If they win the first contract, the probability of winning the second is 70%. However, if they lose the first contract, the president thinks that the probability of winning the second contract decreases to 50%.

a) What is the probability that they win both contracts?
b) What is the probability that they lose both?
c) what is the probability that they win only one contract?

a) do I add the two chances 70+40 then divide it by 200?
b) same as A but with the losing percentages?
c)

Can someone help me with these ? Thank you.

Imagine the bidding process for the two contracts is repeated 100 times.

In about 40 cases they win the first contract, and in 70% of these cases
they win the second, which is 28 cases. So the probability is 28/100=0.28
or about 28%.

So you see that the probability of winning both contracts is:

P(both)=P(first)P(second|first)

which reads: the probability of winning both is the probability of winnimg the
first times the probability that the second is won given that the first has
been won.

Then similarly:

P(none)=P(lose first)P(lose second|lose first)

and:

P(one only)=P(first)P(lose second|first)+P(lose first)P(second|lose first)

(these add because they are mutualy exclusive events)

RonL