1.  Which of these numbers cannot be a probability?


a) -0.00001
0.5
1.001
1


Answer

 Option

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

 Option

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

 Option

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

 Option

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

 Option

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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