• #### 1. In a throw of coin what is the probability of getting head.

1. 1
2. 2
3. 1/2
4. 0

Explanation:

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

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

1. 1
2. 2
3. 1/2
4. 0

Explanation:

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

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

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

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

• #### 4. Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?

1. 1/3
2. 1/6
3. 1/2
4. 1/8

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

• #### 5. In a throw of dice what is the probability of getting number greater than 5

1. 1/2
2. 1/3
3. 1/5
4. 1/6

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

• #### 6. What is the probability of getting a sum 9 from two throws of dice.

1. 1/3
2. 1/9
3. 1/12
4. 2/9

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

1. 3/4
2. 1/4
3. 7/4
4. 1/2