1. Which of these numbers cannot be a probability?

A. a) -0.00001 B. 0.5 C. 1.001 D. 1

Answer Answer: Option D Explanation: A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1.

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

A. 3/1 B. 3/2 C. 3/3 D. 3/4

Answer Answer: Option D Explanation: Here S = {HH,HT, TH,TT}
Let E = event of getting at most one head.
Therefore E = {TT, HT, TH}
Therefore P(E) = n(E)/n(S) = 3/4

3. Two dice are rolled, find the probability that the sum is?

A. equal to 1 B. equal to 4 C. less than 13 D. None of these

Answer Answer: Option A Explanation: a) The sample space S of two dice is shown below.
S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }
Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, hence
P(E) = n(E) / n(S) = 0 / 36 = 0
b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.
P(E) = n(E) / n(S) = 3 / 36 = 1 / 12
c) All possible outcomes, E = S, give a sum less than 13, hence.
P(E) = n(E) / n(S) = 36 / 36 = 1

4. Two coins are tossed, find the probability that two heads are obtained?
{Note: Each coin has two possible outcomes H (heads) and T (Tails)}.

A. 1/2 B. 1/3 C. 1/5 D. 1/4

Answer Answer: Option D Explanation: The sample space S is given by.
S = {(H,T),(H,H),(T,H),(T,T)}
Let E be the event "two heads are obtained".
E = {(H,H)}
We use the formula of the classical probability.
P(E) = n(E) / n(S) = 1 / 4

5. Two coins are tossed, find the probability that two heads are obtained?

A. 1/9 B. 1/4 C. 1/6 D. 1/8

Answer Answer: Option B Explanation: The sample space S is given by.
S = {(H,T),(H,H),(T,H),(T,T)}
Let E be the event "two heads are obtained".
E = {(H,H)}
We use the formula of the classical probability.
P(E) = n(E) / n(S) = 1 / 4