A person draws out two balls successively from a bag containing 6 red and 4 white balls. The probability that at least one of them will be red is  

Let S is sample space.  

$\therefore n\left(S\right){=}^{10}{C}_1{\times }^9{C}_1=90$

Let $E_1$ be the event that I ball drawn is red and II ball drawn is white.

$\therefore n\left(E_1\right){=}^6{C}_1{\times }^4{C}_1=24$

Let $E_2$ be the event that I ball drawn is white and II ball drawn is red

$\therefore n\left(E_2\right){=}^4{C}_1{\times }^6{C}_1=24$

$\therefore$Let $E_3$ be the ever it that both balls drawn are red

$\therefore n\left(E_3\right){=}^6{C}_1{\times }^5{C}_1=30$

$\because E_1,E_2\text{ and }{E}_3\text{ are mutually exclusive. }$

$\therefore n\left(E_1\cup E_2\cup E_3\right)=n\left(E_1\right)+n\left(E_2\right)+n\left(E_3\right)$

                                 $=24+24+30=78$

$\therefore$Probability that at least one of them will be red 

$=\frac{n\left(E_1\cup E_2\cup E_3\right)}{n\left(S\right)}=\frac{78}{90}$