Question

(a) State and prove Addition theorem on probability
(b) Find the probability of drawing an ace or a spade from a well shuffled pack of $52$.

Answer 1

• (a) Addition theorem of probability $\to$ If $A$ and $B$ are any two events then the probability of happening of at least one of the events is defined as
  

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

Proof:-

From set theory, we know that,

$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$

Dividing the above equation by $n\left(S\right)$ both sides we have

$\frac{n(A\cup B)}{n\left(S\right)}=\frac{n\left(A\right)}{n\left(S\right)}+\frac{n\left(B\right)}{n\left(S\right)}-\frac{n(A\cap B)}{n\left(S\right)}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)(\because P\left(X\right)=\frac{n\left(X\right)}{n\left(S\right)})$

Hence proved.

• (b)Total no. of cards, $n\left(S\right)=52$

Total no. of ace, $n\left(X\right)=4$

Total no. of spade $n\left(Y\right)=13$
Probability of withdrawing an ace or spade-

$P=P\left(X\right)+P\left(Y\right)$

$\Rightarrow P=\frac{n\left(X\right)}{n\left(S\right)}+\frac{n\left(Y\right)}{n\left(S\right)}$

$\Rightarrow P=\frac{{}^4{C}_1+{}^{13}{C}_1-1}{{}^{52}{C}_1}=\frac{16}{52}=\frac{4}{13}$

Hence the probability of drawing an ace or a spade from a well shuffled pack of $52$ cards is $\frac{4}{13}$.