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s1: probability stuck on question

Hi, I'm not struggling with a question from my textbook, the answer is P(B=0.5 which works but I don't know how to get it

The only way I could think about doing it would be trial and error which obviously isn't going to get me a good answer!

I was thinking you could do 0.6xP(B) + 0.4xP(B) + 0.6xP(B') =0.7 but that wouldn't work would it as how can we assume P(B)=P(B')?
Reply 1
Original post by Jitesh
Hi, I'm not struggling with a question from my textbook, the answer is P(B=0.5 which works but I don't know how to get it

The only way I could think about doing it would be trial and error which obviously isn't going to get me a good answer!

I was thinking you could do 0.6xP(B) + 0.4xP(B) + 0.6xP(B') =0.7 but that wouldn't work would it as how can we assume P(B)=P(B')?

I'm assuming this is Q12:

Use the formula : P(A or B or both)=P(A)+P(B)P(A and B)P(\text{A or B or both}) = P(A) + P(B) - P(\text{A and B})

Usually written with symbols : P(AB)=P(A)+P(B)P(AB)P(A\cup B) = P(A) + P(B) - P(A \cap B)


Also, since A and B are independent you have the formula

P(A and B)=P(A)×P(B)P(\text{A and B}) = P(A) \times P(B)
Reply 2
Original post by notnek
I'm assuming this is Q12:

Use the formula : P(A or B or both)=P(A)+P(B)P(A and B)P(\text{A or B or both}) = P(A) + P(B) - P(\text{A and B})

Usually written with symbols : P(AB)=P(A)+P(B)P(AB)P(A\cup B) = P(A) + P(B) - P(A \cap B)


Also, since A and B are independent you have the formula

P(A and B)=P(A)×P(B)P(\text{A and B}) = P(A) \times P(B)


Sorry I meant q11, but it seeems like you answered q11? :smile:

Sorry what does the cup mean? and/or ?
Reply 3
Original post by Jitesh
Sorry I meant q11, but it seeems like you answered q11? :smile:

Sorry what does the cup mean? and/or ?

Yes I meant Q11 :smile:

=or\cup = \text{or}

=and\cap = \text{and}
(edited 8 years ago)
Reply 4
Original post by notnek
Yes I meant Q11 :smile:

=or\cup = \text{or}

=and\cap = \text{and}


Sorry I'm confused on to how you would get P(AnB) without knowing the value of B?
Reply 5
Original post by Jitesh
Sorry I'm confused on to how you would get P(AnB) without knowing the value of B?

If you combine the first formula with the second formula I gave you:

P(AB)=P(A)+P(B)P(A)×P(B)P(A\cup B) = P(A) + P(B) - P(A)\times P(B)


You know P(AB)P(A\cup B) and P(A)P(A) so you can work out P(B)P(B).
Reply 6
Original post by notnek
If you combine the first formula with the second formula I gave you:

P(AB)=P(A)+P(B)P(A)×P(B)P(A\cup B) = P(A) + P(B) - P(A)\times P(B)


You know P(AB)P(A\cup B) and P(A)P(A) so you can work out P(B)P(B).


Oh, that's great! Thank you very much!

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