Conditional probability of events A and B `P(A\mid B) = \frac{P(A\cap B)}{P(B)}`

Bayes Theorem

```
\begin{aligned}
P(A \mid B) &= \frac{P(A\cap B)}{P(B)}, \text{ if } P(B) \neq 0, \\
P(B \mid A) &= \frac{P(B\cap A)}{P(A)}, \text{ if } P(A) \neq 0, \\
\Rightarrow P(A\cap B) &= P(A\mid B)\times P(B)=P(B\mid A)\times P(A), \\
\Rightarrow P(A \mid B) &= \frac{P(B \mid A) \times P(A)} {P(B)}, \text{ if } P(B) \neq 0.
\end{aligned}
```

**A couple has two children, the older of which is a boy. What is the probability that they have two boys?**

```
\begin{aligned}
P(B) &= Older one is boy \\
P(A) &= Two boys \\
\\
\\
P(A|B) &= \frac{P(B/A) P(A)}{P(B)} \\
&=\frac{P(older~one~is~boy~given~two~boys) * P(having~two~boys)}{P(older~one~is~boy)} \\
Given Fact, \\
P(older~one~is~boy~given~two~boys) &= 1 \\
P(A|B) &= \frac{1 * 1/2*1/2}{1/2} \\
&= 1/2
\end{aligned}
```

**A couple has two children, one of which is a boy. What is the probability that they have two boys?**

```
\begin{aligned}
A &= having two boys\\
B &= one child is a boy\\
\\
\\
P(one~child~boy) &= 1-P(both~children~not~boys)\\
&= 1- 1/4 \\
&= 3/4\\
P(having~two~boys~given~one~child~boy) &= \frac{P(one~child~is~boy~gicen~both~children~boys) * P(both~children~boys)}{P(one~child~boy)} \\
&=\frac{1 * 1/4}{3/4} \\
&= 1/3\\
\end{aligned}
```

**A family has two children. Given that one of the children is a boy, and that he was born on a Tuesday, what is the probability that both children are boys?**

```
\begin{aligned}
P(BB) &= Probablity~of~having~both~boys~of~two~children \\
P(B) &= Probablity~of~having~one~boy~of~two~children\\
P(BT) &= probablity~of~having~a~boy~born~on~Tuesday \\
\\
P(BB|BT) &= \frac{P(BT|BB) *P(BB)}{ P(BT)} \\
P(BT|BB) &=P(having~boy~born~on~Tuesday~given~both~children~are~boys) \\
P(BT|BB) &=P(Either~of~two~can~be~boy~~born~on~Tuesday~given~both~boys) \\
\\
P(BT|BB) &= 1- P(boys~born~on~other~days)\\
&=1-6/7*6/7 \\
&= 1-36/49 = 13/49\\
\\
P(BB) &= 1/2*1/2 = 1/4 \\
\\
P(BT) &= P(one~of~the~child~is~a~boy~born~on~Tuesday) \\
&=1-P((both~boys~born~on~non-Tuesdays) + both~girl) \\
&= 1- 13/14 * 13/14 \\
&= 1- 169 /196 = 27/196\\
\\
P(BB|BT) &= P(BT|BB) *P(BB)/ P(BT) \\
&= 13/49 * 1/4 / 27/196 \\
&=13/49 * 1/4 * 196 /27 \\
&= 13/27
\end{aligned}
```

**Balls numbered 1 through 20 are placed in a bag. Three balls are drawn out of the bag without replacement. What is the probability that all the balls have odd numbers on them**

Three balls drawn with out replacement

Total Probability = P(1st ball) * P(2nd ball) * P(3rd ball)

**= 10/20 9/19 8/18**