Probability

[Dr Finlay]

I decided to go over the probability section in my S1 statistics book. However, it seems i've stumbled at the first hurdle. The first exercise is about the addition rule P(AUB) = P(A) + P(B) - P(AnB). There is a question where it asks about the probabilities events not happening, P(A'), P(A'UB), P(A'nB') etc. I know that P(A') = 1 - P(A), but i'm completely stuck on how to get such probabilities of P(A'nB).

 

Can anyone help unclog my mental block?

 

Thanks once again,

Rob

[Primarygun]

[Math] P(A\cap B) [/Math]

refers to the probability of the event that is favourable to both conditions=P(A and B)

Say, A refers to an odd no. and B refers to prime no.

A=(1,3,5) and B=(2,3,5)

Then, for a dice, favourable outcomes for that P are (3,5)

[Dr Finlay]

I managed to get that far, but here's an example of a question.

 

If S and T are two events and P(T) = 0.4 and P(SnT) = 0.15 and P(S' n T') = 0.5, find:

 

a) P(S n T')

b) P(S)

c) P(S u T)

d) P(S' n T)

e) P(S' u T')

 

All these questions should be able to be solved using the addition rule. But I can't figure out how to get probabilities of events not occurring from events that do occur, such as a), d) and e).

[Dr Finlay]

Just done another question, i get a different answer from the book though.

 

If A and B are two events and P(A) = 0.6 and P(B) = 0.3 and P(AuB) = 0.8 find:-

a) P(AnB)

b)P(A'nB)

c)P(AnB')

 

For part a) i got 0.1 using the addition rule, which the book says is correct. But for part b) i did

P(A'nB) = (1 - P(A))xP(B)

= 0.4 x 0.3

= 0.12

 

the book however says the answer is 0.2.

 

For part c) i did:

 

P(AnB') = P(A) x (1 - P(B))

= 0.6 x 0.7

= 0.42

 

the book gives the answer 0.5

 

Can anyone see what I'm doing wrong?

 

Thanks alot,

Rob

[Dr Finlay]

Hmm, i think i may have worked it out. I drew a Venn diagram and got

 

[math]P(A \cap B) = P(B) - P(A \cap B)[/math]

 

so it follows

 

[math]P(A \cap B') = P(A) - P(A \cap B)[/math]

 

i then get the book's answers.

I think i just needed to spend the time to work it out, cheers anyway.

[Zone Ranger]

Hmm' date=' i think i may have worked it out. I drew a Venn diagram and got

 

[math']P(A \cap B) = P(B) - P(A \cap B)[/math]

 

so it follows

 

[math]P(A \cap B') = P(A) - P(A \cap B)[/math]

 

i then get the book's answers.

I think i just needed to spend the time to work it out, cheers anyway.

 

[math]P(A' \cap B) = P(B) - P(A \cap B)[/math]