1.  If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. What is the probability of getting?


three head
two head
one head
0 head


Answer

 Option

Total number of trials = 175. Number of times three heads appeared = 21. Number of times two heads appeared = 56. Number of times one head appeared = 63. Number of times zero head appeared = 35. Let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and zero head respectively. (i) P(getting three heads) Number of times three heads appeared = P(E1) = total number of trials = 21/175 = 0.12 (ii) P(getting two heads) Number of times two heads appeared = P(E2) = total number of trials = 56/175 = 0.32 (iii) P(getting one head) Number of times one head appeared = P(E3) = total number of trials = 63/175 = 0.36 (iv) P(getting zero head) Number of times zero head appeared = P(E4) = total number of trials = 35/175 = 0.20 Note: Remember when 3 coins are tossed randomly, the only possible outcomes are E2, E3, E4 and P(E1) + P(E2) + P(E3) + P(E4) = (0.12 + 0.32 + 0.36 + 0.20) = 1

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2.  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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3.  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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4.  In how many different ways the word FAVOUR can be arranged?


820
530
560
720


Answer

 Option

6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

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5.  The probability of two persons of passing the interview are 1/3 and 3/5. Then calculate the probability that only one of them pass the interview?


7/15
8/15
11/15
13/15


Answer

 Option

PF + FP = (P = pass, F = fail) 1/3 * 2/5 + 2/3*3/5 = 8/15

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