probability help plz

**hasnain721** · December 7th 2007, 11:35 AM

Hi there,
I am doing s1 for this jan and i am finding it very difficult to cope up. Especially for probability. I have a cgp buk but stil its not very gud at probability. Here is a question from my text buk which i cud not understand : -

Two events A and B are such that P(A) = 3/4 , P(B l A) = 1/5 and P(B' l A') = 4/7. By use of a probability tree or other wise find :

a) P(A and B)

b) P(B)

c) P(A l B)

In addition, can some 1 suggest a gud buk for probability plz.

Thanks!

**Soroban** · December 7th 2007, 01:08 PM

Hello, Hasnain!

Two events $A$ and $B$ are such that: . $P(A) \:=\:\frac{3}{4},\;\;P(B|A)\:=\:\frac{1}{5},\;\;P( B'|A') \:= \:\frac{4}{7}$

Find: . $(a)\;P(A \cap B)\qquad (b)\;P(B)\qquad (c)\;P(A|B)$

I assume you're familiar with Bayes' Theorem: . $P(X|Y) \:=\:\frac{P(X \cap Y)}{P(Y)}$

(a) We have: . $P(B|A) \:=\:\frac{P(B \cap A)}{P(A)}$

Since $P(B|A) = \frac{1}{5}$ and $P(A) = \frac{3}{4}$

. . the equation becomes: . $\frac{1}{5} \:=\:\frac{P(A \cap B)}{\frac{3}{4}}\quad\Rightarrow\quad P(A \cap B) \:=\:\frac{1}{5}\cdot\frac{3}{4}$

Therefore: . $\boxed{P(A \cap B) \;=\;\frac{3}{20}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(b) There are a number of ways to determine $P(B).$
Here's one of them . . .

We are given: . $P(B'|A') \:=\:\frac{P(B' \cap A')}{P(A')} \:=\:\frac{4}{7}$

Since $P(A) = \frac{3}{4}$ , then $P(A') = \frac{1}{4}$
. . and the equation becomes: . $\frac{P(A' \cap B')}{\frac{1}{4}} \:=\:\frac{4}{7}\quad\Rightarrow\quad P(A' \cap B') \:=\:\frac{1}{7}$

Sketching a Venn diagram, we see that:

. . $P(A) + P(A \cap B) + P(A' \cap B') + P(A' \cap B) \;=\;1$

. . . $\frac{3}{5}\quad + \quad\; \frac{3}{20}\quad \;\;+\quad\quad \frac{1}{7} \quad+\quad P(A'\cap B) \;=\;1$

Then: . $P(A' \cap B) \:=\:1 - \frac{3}{5} -\frac{3}{20} - \frac{1}{7}\;=\;\frac{3}{28}$

Hence: . $P(B) \;=\;P(A\cap B) + P(A' \cap B) \;=\;\frac{3}{20} + \frac{3}{28}$

Therefore: . $\boxed{P(B)\;=\;\frac{9}{35}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(c) We've done the hard work ... This part is easy . . .

$P(A|B) \;=\;\frac{P(A\cap B)}{P(B)}\;=\;\frac{\dfrac{3}{20}}{\dfrac{9}{35}} \;=\;\boxed{\frac{7}{12}}$

**badgerigar** · December 7th 2007, 11:56 PM

In addition, can some 1 suggest a gud buk for probability plz.

I quite like Gnedenko,s "Theory of Probability".

**hasnain721** · December 8th 2007, 04:10 AM

Hi there,
Thanks a lot for replying. I understand what u've done till you worked out
P( A' ∩ B'). Is that the same as P( B' ∩ A') ?

Secondly, when you said, " Sketching a Venn diagram, we see that:",
i dont understand which part of the diagram is P(A' ∩ B) . If you dont mind, can u sketch it plz?

Thanks a lot for replyin and i have checked the answers at the back of the book and they are right!

Thanks.
Hasnain Mir Mohammed

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