Events A, B independent; P(A) = 0.45, probability not A or B = 0.70; find P(B)

Consider two independent events, A and B, where the P(A) is 0.45 and the probabilitythat A does not occur or B occurs is 0.70. Determine the probability that event B occurs.

it looks easy but it is sure hard for me!!

Please help!!!

kawsarkurban said:

Consider two independent events, A and B, where the P(A) is 0.45 and the probabilitythat A does not occur or B occurs is 0.70. Determine the probability that event B occurs.

it looks easy but it is sure hard for me!!

Please help!!!

Click to expand...

I will have a feeble attempt at this.
P( not A)=0.55 (1-0.45)
P(B not A)=0.70 given

P(B)=P(B not A)-P(not A)=0.7-0.55=0.15

I believe you should be able to add up 0.15 +0.7 to give you both the situations of B occurring when A does and also B occurring when A does not to give you 0.85
Wait for other members to confirm or refute the above because I am not totally sure if it is correct.

m2017 said:

I will have a feeble attempt at this.
P( not A)=0.55 (1-0.45)
P(B not A)=0.70 given

P(B)=P(B not A)-P(not A)=0.7-0.55=0.15

I believe you should be able to add up 0.15 +0.7 to give you both the situations of B occurring when A does and also B occurring when A does not to give you 0.85
Wait for other members to confirm or refute the above because I am not totally sure if it is correct.

Click to expand...

We need to fix your notation; to me "B not A" would mean "B and not A". So here's what you're saying:

P(not A)= 1-0.45 = 0.55
P(B or not A)=0.70

P(B)=P(B or not A)-P(not A)=0.7-0.55=0.15

Click to expand...

This is wrong because B and not A are not mutually exclusive, so you can't know that P(B) + P(not A) = P(B or not A).

I would start by making a Venn diagram and shading in what B or not A is. If that region is .70, its complement is .30. What region is that complement? (You could also do this by using de Morgan's Laws to find the complement of B or not A, but I'm guessing neither of you is very deeply into set theory yet.)

Then you'll want to use the fact that A and B are independent.

kawsarkurban said:

Consider two independent events, A and B, where the P(A) is 0.45 and the probabilitythat A does not occur or B occurs is 0.70. Determine the probability that event B occurs.

it looks easy but it is sure hard for me!!

Please help!!!

Click to expand...

Consider two independent events, A and B. So P(A and B) = P(A)P(B)

probabilitythat A does not occur or B occurs is 0.70. P(not A or B)=.70

You should know that since A and B are independent then so are not A and not B, not A and B and A and not B.

See what you can do from here.

Jomo said:

Consider two independent events, A and B. So P(A and B) = P(A)P(B)
You should know that since A and B are independent then so are not A and not B, not A and B and A and not B.

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Here is a quick proof that last point.
\(\displaystyle \begin{align*} P(B \cap \neg A) &= P(B) - P(B \cap A)\\ &= P(B) - P(B)P(A)\\ &= P(B)\left[ {1 - P(A)} \right]\\ &= P(B)P(\neg A) \end{align*}\)

Do you see how that proves that \(\displaystyle B~\&~\neg A\) are independent?

If \(\displaystyle B~\&~A\) are independent then \(\displaystyle B~\&~\neg A\) are independent.
If \(\displaystyle \neg A~\&~B\) are independent then \(\displaystyle \neg A~\&~\neg B\) are independent.
etc.

pka said:

Here is a quick proof that last point.
\(\displaystyle \begin{align*} P(B \cap \neg A) &= P(B) + P(B \cap A)\\ &= P(B) + P(B)P(A)\\ &= P(B)\left[ {1 - P(A)} \right]\\ &= P(B)P(\neg A) \end{align*}\)

Do you see how that proves that \(\displaystyle B~\&~\neg A\) are independent?

If \(\displaystyle B~\&~A\) are independent then \(\displaystyle B~\&~\neg A\) are independent.
If \(\displaystyle \neg A~\&~B\) are independent then \(\displaystyle \neg A~\&~\neg B\) are independent.
etc.

Click to expand...

You had me going there a bit until I realized that your + signs really should be - signs. Nice proof!

Jomo said:

You had me going there a bit until I realized that your + signs really should be - signs. Nice proof!

Click to expand...

Thanks for the correction.