In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?

Answer: Option D

Explanation:

Total 4 cases = [HH, TT, TH, HT]
Favourable cases = [HH, TH, HT]
Please note we need atmost one tail, not atleast one tail.

So probability = 3/4

Answer: Option C

Explanation:

Total cases = [H,T] - 2
Favourable cases = [H] -1
So probability of getting head = 1/2

Answer: Option C

Explanation:

\begin{aligned}
\text{Total cases =} ^{52}C_2 \\
= \frac{52*51}{2*1} = 1326 \\
\text{Total King cases =} ^{4}C_2 \\
= \frac{4*3}{2*1} = 6 \\

\text{Probability =} = \frac{6}{1326}\\
= \frac{1}{221} \\
\end{aligned}

Answer: Option C

Explanation:

\begin{aligned}
\text{Total cases =} ^{12}C_3 \\
= \frac{12*11*10}{3*2*1} = 220 \\
\text{Total cases of drawing same colour =} \\
^{5}C_3 + ^{4}C_3 + ^{3}C_3 \\
\frac{5*4}{2*1} + 4 + 1 = 15 \\

\text{Probability of same colur =} = \frac{15}{220}\\
= \frac{3}{44} \\

\text{Probability of not same colur =} \\
1-\frac{3}{44}\\ = \frac{41}{44}

\end{aligned}

Answer: Option C

Explanation:

Total number of cases = 52
Favourable cases = 2

Probability = 2/56 = 1/26