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

Answer: Option D

Explanation:

Number greater than 5 is 6, so only 1 number
Total cases of dice = [1,2,3,4,5,6]

So probability = 1/6

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 B

Explanation:

Total number of cases = 6*6 = 36

Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4

So probability = 4/36 = 1/9

Answer: Option C

Explanation:

Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]

Favoured cases are = [TTH, THT, HTT, TTT] = 4

So required probability = 4/8 = 1/2

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