1.  Which of these numbers cannot be a probability?

a) -0.00001
0.5
1.001
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.
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2.  Two unbiased coins are tossed. What is the probability of getting at most one head?

3/1
3/2
3/3
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 ```

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

equal to 1
equal to 4
less than 13
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
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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)}.

1/2
1/3
1/5
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  ```

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

1/9
1/4
1/6
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  ```

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