Question Detail

In a throw of coin what is the probability of getting tails.

Answer: Option C

Explanation:

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

1. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?

Answer: Option A

Explanation:

Total number of cases = 6*6 = 36

Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27

So Probability = 27/36 = 3/4

2. Two unbiased coins are tossed. What is probability of getting at most one tail ?

\begin{aligned} \frac{1}{2} \end{aligned}
\begin{aligned} \frac{1}{3} \end{aligned}
\begin{aligned} \frac{3}{2} \end{aligned}
\begin{aligned} \frac{3}{4} \end{aligned}

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

3. From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings

2/121
2/221
1/221
1/13

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}

4. 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?

2/3
8/21
3/7
9/22

Answer: Option B

Explanation:

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.

5. In a throw of coin what is the probability of getting head.

Answer: Option C

Explanation:

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