Conditional Probability

[Aladdin]

Good Morning Free Math Help Forum !!

Question # 1)
If P(A) = 0.8 , P(B/A)= 0.3 and P(B/A(Bar))=0.2 then P(A) = ??
The Respnses are : 0.5 or 0.28 or 0.24 .

Question # 2)
The coefficient of (x^5)(y^2) in the expanding of (2x - 3y)^7 is :
The Respnses are : 6048 or (Combination 7)^2(2^5) or (Combination 7)^2(3^2)
Sorry I don't know how to type the symbol of combination correct.

My Work ::

Q1) P(B/A) = P(BnA)/P(A) --> P(BnA) = (0.8)(0.3) = 0.24

P(A(bar)) = 1 - P(A) = 0.2 --> P(BnA(bar) = 0.2*0.2 = 0.4

P(B) = P(BnA) + P(BnA(bar) !!
P(B0 = 0.64 which doesn't exist in the respnses ::

Q2) Binomial Theorem :
(C7^p)(a^(7-p))(b^p)

By comparison I got 7- p = 5 , p = 2
So the coefficient is C7^2
Which also dosn't exist !!

Thanks in advance.
Aladdin

 

[galactus]

Coefficient of $\displaystyle x^{5}y^{2}$ in the expansion of $\displaystyle (2x-3y)^{7}$

The binomial expansion has a pattern:

$\displaystyle (a+b)^{7}=\binom{7}{0}a^{7}+\binom{7}{1}a^{6}b+\binom{7}{2}a^{5}b^{2}+....................+\binom{7}{7}b^{7}$

Plug in $\displaystyle a=2x \;\ b=-3y$

Then, the term we are looking for becomes:

$\displaystyle (2x)^{5}(-3y)^{2}=288x^{5}y^{2}$

But note the coefficient is $\displaystyle \binom{7}{2}=21$

So, we get $\displaystyle \fbox{6048}x^{5}y^{2}$


For the first one, you have P(A)=.8 and then P(A)=?. I assume there is a typo?.

 

[soroban]

Hello, Aladdin!

As galactus pointed out, there is a typo in #1 . . .

$\displaystyle (1) \;\text{Given: }\;P(A) \,=\, 0.8,\;\; P(B|A)\,=\, 0.3,\;\;P(B|A')\,=\,0.2$

. . .\(\displaystyle \text{Find: }\(B)\)

$\displaystyle \text{The responses are: }\;(a)\;0.5\quad (b)\;0.28\quad (c)\;0.24$

\(\displaystyle \text{We are given: }\(A) \,=\,0.8 \quad\Rightarrow\quad P(A') \,=\,0.2\)

$\displaystyle \text{We will use Bayes' Theorem repeatedly.}$


\(\displaystyle \text{We have: }\(B|A) \:=\:0.3 \quad\Rightarrow\quad \frac{P(B \wedge A)}{P(A)} \:=\:0.3 \quad\Rightarrow\quad \frac{P(B \wedge A)}{0.8} \:=\:0.3 \quad\Rightarrow\quad P(B\;\wedge\;A) \:=\:0.24\)

\(\displaystyle \text{We have: }\(B|A') \:=\:0.2 \quad\Rightarrow\quad \frac{P(B \wedge A')}{P(A')} \:=\:0.2 \quad\Rightarrow\quad \frac{P(B \wedge A')}{0.2} \:=\:0.2 \quad\Rightarrow\quad P(B\;\wedge\;A') \:=\:0.04\)


\(\displaystyle \text{Therefore: }\(B) \;=\;P(B \wedge A) + P(B \wedge A') \;=\;0.24 + 0.04 \;=\;\boxed{0.28}\)

 

[Aladdin]

Thanks for the help guys : :

I really appreciate it.

0.2*0.2 = 0.04 :wink:

 

[singaravelu]

if a and b are mutually exclusive events and p(avb)not equall to zero,prove that p(a/a union b)=p(a)/p(a)+p(b)

 

[singaravelu]

if a and b are mutually exclusive events and p(a union b)not equall to zero,prove that p(a/a union b)=p(a)/p(a)+p(b)

 

[soroban]

Hello, singaravelu!

$\displaystyle \text{if }a\text{ and }b\text{ are mutually exclusive events, and }P(a\cup b)\,\neq\,0$

. . $\displaystyle \text{prove that: }\;P(a\,|\,a \cup b) \:=\:\frac{P(a)}{P(a)+P(b)}$

$\displaystyle \text{Bayes' Theorem: }\;P(a\,|\,a\cup b) \;=\;\frac{P(a \cap (a\cup b))}{P(a\cup b)}\;\;[1]$


\(\displaystyle \text{The numerator is: }\(a \cap (a \cup b)) \:=\(a)\)

\(\displaystyle \text{The denominator is: }\(a \cup b) \:=\(a) + P(b)\)


$\displaystyle \text{Substitute into [1]: }\;P(a\,|\,a\cup b) \;=\;\frac{P(a)}{P(a)+P(b)}$