1. A coin and a die are thrown once, then the sample space will be of ……….. ordered pairs. Easy
A. 2*6
B. 1*5
C. 2*12
D. None of these

View Answer

Answer: Option A

Explanation:

If H and T represent head and tail of the coin which can appear when the coin falls and 1, 2, 3 ….., 6 are the numbers which can appear on the upper face of the die when it falls, then the required sample space will be the following set S of 2*6 ordered pairs.

S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

2. In the throw of a die, find the probability of obtaining a number greater than 2. Easy
A. 1/3
B. 2/3
C. 1/6
D. None of these

View Answer

Answer: Option B

Explanation:

The numbers on a die greater than 2 are four viz. 3, 4, 5, 6 and numbers not greater than 2 are two viz, 1, 2
Therefore, the required probability = n(E)/n(S) = 4/(4+2) = 2/3
3. Two dice are thrown simultaneously. What is the probability of obtaining a total score of seven? Easy
A. 1/7
B. 2/6
C. 1/12
D. None of these

View Answer

Answer: Option D

Explanation:

Numbers 1 to 6 (one at a time) are marked on the six faces of each die.
-> There are six possible ways of getting a number on the upper face of one die and corresponding to each of these six ways, there are six possible numbers on the upper face of the second die.
Total number of ways of getting numbers on the upper faces of the two dice = 6*6 = 36 = n(S), where S is the sample space.
Let E be the event of obtaining a total score of 7, then
E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, where (1,6) means number 1 on the first die and number 6 on the second.
Therefore n(E) = 6
Required probability = n(E)/n(S) = 6/36 = 1/6
4. In a single throw with two dice, what is the chance of throwing two aces? Easy
A. 1/12
B. 1/18
C. 1/36
D. None of these

View Answer

Answer: Option C

Explanation:

If S be the sample space, then n(S) = total number of exhaustive cases = 6*6 = 36
Also, two aces i.e. (1,1) can be thrown in one way only, so if E be this event, then n(E) = 1
Required probability = n(E)/n(S) = 1/36
5. From a pack of 52 cards, four cards are drawn. Find the chance that they will be the four honours of the same suit. Medium
A. 2/270725
B. 4/270725
C. 1/270725
D. None of these

View Answer

Answer: Option B

Explanation:

Here, number of exhaustive cases = 52C4­

There are four suits and each suit contains four honours (ace, king, queen and knave), so the number of favourable cases = number of ways in which one suit of 4 honours can be selected from these 4 suits of 4 honours each = 4C1

Required probability = (number of favourable cases)/(number of exhaustive cases) = 4C1/52C4 = ((4)*4*3*2*1)/(52*51*50*49) = 4/270725

Note: Number of possible combinations of r objects from a set of n objects is written as:

6. What is the chance that a leap year, selected at random will contain 53 Sundays? Medium
A. 1/7
B. 3/7
C. 2/7
D. None of these

View Answer

Answer: Option C

Explanation:

Number of days in a leap year is 366 and so it contains 52 complete weeks and two days over. These two days may be one of the following seven combinations:
i. Monday and Tuesday
ii. Tuesday and Wednesday
iii. Wednesday and Thursday
iv. Thursday and Friday
v. Friday and Saturday
vi. Saturday and Sunday
vii. Sunday and Monday
Out of these seven equally likely cases the last two contain Sunday and so number of favourable cases is two and the unfavourable is 7 -2 i.e. 5.
Therefore, number of exhaustive cases = 2+5 = 7
So the required probability = = (number of favourable cases)/(number of exhaustive cases) = 2/7

7. What is the chance that a non-leap year should have 53 Sundays? Easy
A. 1/7
B. 3/7
C. 2/7
D. None of these

View Answer

Answer: Option A

Explanation:

Number of days in a non-leap year is 365 and so it contains 52 complete weeks and one day over. This one day over may be one of the seven days of the week viz. Sunday, Monday, Tuesday …., Saturday
Therefore, number of favourable cases = 1 and number of unfavourable cases = 7-1 = 6
Number of exhaustive cases = 1+6 = 7
Required probability = (number of favourable cases)/(number of exhaustive cases) =1/7

8. A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning this bet? Medium
A. 4 to 9
B. 9 to 4
C. 4 to 13
D. None of these

View Answer

Answer: Option B

Explanation:

Total numbers of cards in an ordinary pack is 52. Out of these 13 are spades and 4 are aces (These four aces include one of spade also).
Therefore, number of exhaustive cases = number of ways in which a card can be spade or an ace = 13+(4-1) = 16
The probability of the gambler’s winning the bet = 16/52 = 4/13 = 4/(4+9)
Required odds against his winning the bet are 9 to 4.

9. Two dice are thrown. What is the probability that the product of the numbers on their uppermost faces lies between 7 and 13? Difficult
A. 1/4
B. 3/4
C. 1/9
D. None of these

View Answer

Answer: Option A

Explanation:

Here the sample space S contains 6*6 i.e. 36 elements and
so, n(S) = 36
Also, if E be the event that product of the numbers appearing on the uppermost faces of the dice lies between 7 and 13, then
E = {(2,4), (2,5), (2,6), (3,3), (3,4), (4,2), (4,3), (5,2), (6,2)}
-> n(E) = 9
-> Required probability = n(E)/n(S) = 9/36 = 1/4

10. A bag contains 3 black and 4 white balls. What is the probability of drawing a white ball? Medium
A. 3/7
B. 4/7
C. 5/7
D. None of these

View Answer

Answer: Option B

Explanation:

Total number of balls in the bags = 3+4 = 7
n(S) = number of elements in the sample space = number of ways in which one ball can be drawn from the bag = 7C1 = 7
Also, E if be the event of drawing a white ball, then n (E) = number of ways in which one white ball can be drawn out of four = 4C1 = 4
Required probability = n(E)/n(S) = 4/7