# 10 - Maths - Unit 12. Probability

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Prepared by
Mr. A. GANESAN, VICE PRINCIPAL, SHRI GURU BRAMMA VIDHYALAYA, KARUR-5
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• 1.

### If S is the sample space of a random experiment, then P(S) =

• A.

1

• B.

0

• C.

1/8

• D.

1/2

A. 1
Explanation
The probability of an event occurring in a sample space is always 1, as the sample space represents all possible outcomes of the random experiment. Therefore, the probability of the sample space itself occurring is 1.

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• 2.

### If p is probability of an event A, then p satisfies =

• A.

0 ≤ p ≤ 1

• B.

0 < p < 1

• C.

0 ≤ p < 1

• D.

0 < p ≤ 1

A. 0 ≤ p ≤ 1
Explanation
The probability of an event A, denoted as p, must satisfy the condition that it is greater than or equal to 0 and less than or equal to 1. This is because the probability of an event can range from 0 (indicating impossibility) to 1 (indicating certainty). Therefore, the correct answer is 0 ≤ p ≤ 1.

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• 3.

### Let A and B be any two events and S be the corresponding sample space. Then P(Ā ∩B)=

• A.

P(B)- P(A∩B)

• B.

P (A ∩B)- P(B)

• C.

P(S)

• D.

P[ (AUB)']

A. P(B)- P(A∩B)
Explanation
The given answer, P(B)- P(A∩B), is the correct formula for calculating the probability of the complement of event A intersecting with event B. This formula subtracts the probability of A and B occurring together from the probability of event B occurring on its own. This is because the intersection of A and B is included in both events A and B, so we need to subtract the overlapping probability to avoid double counting.

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• 4.

### The Probability that a student will score centum in mathematics is 4/5. The Probability that he will not score centum is =

• A.

1/5

• B.

2/5

• C.

3/5

• D.

4/5

A. 1/5
Explanation
The probability that a student will score centum in mathematics is 4/5. This means that there is a 4/5 chance that the student will achieve a perfect score. The probability that he will not score centum is therefore the complement of this, which is 1 - 4/5 = 1/5. So, the probability that he will not score centum is 1/5.

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• 5.

### If A and B are two events such that P(A) = 0.25, P(B) = 0.05 and P(A∩B)=0.14, then P(AUB) =

• A.

0.16

• B.

0.61

• C.

0.14

• D.

0.6

A. 0.16
Explanation
The probability of the union of two events, P(AUB), can be calculated by adding the probabilities of the individual events (A and B) and then subtracting the probability of their intersection (A∩B). In this case, P(A) = 0.25, P(B) = 0.05, and P(A∩B) = 0.14. Therefore, P(AUB) = P(A) + P(B) - P(A∩B) = 0.25 + 0.05 - 0.14 = 0.16.

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• 6.

### There are 6 defective items in a sample of 20 items. One item is drawn at random. The Probability that it is a non-defective item is

• A.

7/10

• B.

0

• C.

3/10

• D.

2/3

A. 7/10
Explanation
In a sample of 20 items, there are 6 defective items. Therefore, there are 20 - 6 = 14 non-defective items. The probability of drawing a non-defective item is the number of favorable outcomes (14) divided by the total number of possible outcomes (20). Thus, the probability is 14/20, which simplifies to 7/10.

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• 7.

### If A and B are mutually exclusive events and S is the Sample space such that  P(A) =1/3P(B) and S= AUB, then P(A)=

• A.

1/4

• B.

1/2

• C.

3/4

• D.

3/8

A. 1/4
Explanation
If A and B are mutually exclusive events, it means that they cannot occur at the same time. Therefore, the probability of the union of A and B (S=AUB) is equal to the sum of the probabilities of A and B.

Given that P(A) = 1/3P(B), we can substitute this into the equation P(A) + P(B) = P(AUB), which gives us P(A) + 3P(A) = 1.

Simplifying the equation, we get 4P(A) = 1, and dividing both sides by 4, we find that P(A) = 1/4.

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• 8.

### The Probabilities of three mutually exclusive events A, B  and C are given by 1/3,1/4 and 5/12. Then P(AUBUC)  is

• A.

1

• B.

19/12

• C.

11/12

• D.

7/12

A. 1
Explanation
The probability of the union of three mutually exclusive events A, B, and C is equal to the sum of their individual probabilities. Since the events A, B, and C are mutually exclusive, they cannot occur at the same time. Therefore, the probability of their union is equal to the sum of their probabilities, which is 1/3 + 1/4 + 5/12 = 12/36 + 9/36 + 15/36 = 36/36 = 1.

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• 9.

### If P(A)= 0.25 , P(B)= 0.50 , P(AB) = 0.14 then  P(neither A or B ) =

• A.

0.39

• B.

0.25

• C.

0.11

• D.

0.24

A. 0.39
Explanation
The probability of neither A nor B occurring can be calculated by subtracting the probability of A or B occurring from 1. Since P(A) = 0.25 and P(B) = 0.50, the probability of A or B occurring is P(A) + P(B) - P(AB) = 0.25 + 0.50 - 0.14 = 0.61. Therefore, the probability of neither A nor B occurring is 1 - 0.61 = 0.39.

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• 10.

### A bag contains 5 black balls , 4 white balls and 3 red balls. If a ball is selected at random , the probability that it is not red is =

• A.

3/4

• B.

5/12

• C.

4/12

• D.

3/12

A. 3/4
Explanation
The probability that a ball is not red can be found by adding the probabilities of selecting a black ball and a white ball. The probability of selecting a black ball is 5/12 and the probability of selecting a white ball is 4/12. Adding these probabilities gives us 9/12, which simplifies to 3/4.

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• 11.

### Two dice are thrown simultaneously. The probability of getting a doublet is =

• A.

1/6

• B.

1/36

• C.

1/3

• D.

2/3

A. 1/6
Explanation
When two dice are thrown simultaneously, there are a total of 36 possible outcomes (6 possible outcomes for the first dice multiplied by 6 possible outcomes for the second dice). Out of these 36 outcomes, there are 6 outcomes where a doublet is obtained (1-1, 2-2, 3-3, 4-4, 5-5, 6-6). Therefore, the probability of getting a doublet is 6/36, which simplifies to 1/6.

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• 12.

### A fair die is thrown once . The probability of getting a prime or composite number is

• A.

5/6

• B.

1

• C.

0

• D.

1/6

A. 5/6
Explanation
When a fair die is thrown once, it can land on any number from 1 to 6 with equal probability. Out of these numbers, only 2, 3, and 5 are prime numbers, while 1, 4, and 6 are composite numbers. Therefore, the probability of getting a prime or composite number is 5 out of 6, which can be represented as 5/6.

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• 13.

### Probability of getting 3 heads or 3 tails in tossing a coin 3 times is

• A.

1/4

• B.

1/8

• C.

3/8

• D.

1/2

A. 1/4
Explanation
The probability of getting 3 heads or 3 tails in tossing a coin 3 times can be calculated by considering the different possible outcomes. In this case, there are only two possible outcomes that satisfy the condition: getting either 3 heads or 3 tails. Out of the total 8 possible outcomes (2^3), only 2 outcomes meet the condition. Therefore, the probability is 2/8, which simplifies to 1/4.

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• 14.

### A card is drawn from a pack of 52 cards at random. The probability of getting neither an ace  nor a king card is

• A.

11/13

• B.

2/13

• C.

4/13

• D.

8/13

A. 11/13
Explanation
When drawing a card from a pack of 52 cards, there are 4 aces and 4 kings in the deck. Since we want to find the probability of not getting an ace or a king, we need to subtract the probability of getting an ace or a king from 1. The probability of getting an ace is 4/52 and the probability of getting a king is also 4/52. Therefore, the probability of not getting an ace or a king is 1 - (4/52 + 4/52) = 44/52 = 11/13.

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• 15.

### The probability that a leap year will have 53 fridays or 53 saturdays is

• A.

3/7

• B.

2/7

• C.

1/7

• D.

4/7

A. 3/7
Explanation
A leap year has 52 weeks and 2 extra days. These extra days can be any combination of weekdays, including Friday and Saturday. There are 7 possible combinations for these two extra days: Friday-Friday, Friday-Saturday, Saturday-Friday, Saturday-Saturday, Sunday-Saturday, Monday-Saturday, and Tuesday-Saturday. Out of these 7 combinations, 3 include either 53 Fridays or 53 Saturdays. Therefore, the probability is 3/7.

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• 16.

### The Probability that a non-leap year will have 53 Sundays and 53 Mondays is

• A.

0

• B.

1/7

• C.

2/7

• D.

3/7

A. 0
Explanation
A non-leap year has 365 days, which is not divisible by 7. Therefore, each day of the week will occur 52 times in a non-leap year. It is not possible to have 53 Sundays and 53 Mondays in a non-leap year because it would require a total of 106 days, which is more than the total number of days in a year. Hence, the probability is 0.

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• 17.

### The Probability of selecting a queen of hearts when a cards is drawn from a pack of 52 playing cards is

• A.

1/52

• B.

16/52

• C.

1/13

• D.

1/26

A. 1/52
Explanation
The probability of selecting a queen of hearts when a card is drawn from a pack of 52 playing cards is 1/52 because there is only one queen of hearts out of the total 52 cards in the deck. Therefore, the chance of selecting the queen of hearts is 1 out of 52.

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• 18.

### The Probability of sure event is

• A.

1

• B.

0

• C.

100

• D.

0.1

A. 1
Explanation
The probability of a sure event is 1 because it is certain to occur. In probability theory, an event with a probability of 1 is considered to be guaranteed to happen. This means that there is no possibility of it not occurring. Therefore, the correct answer is 1.

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• 19.

### The outcome of a random experiment results in either success or failure. If the probability  of success is twice the probability of failure, then the probability of success is

• A.

2/3

• B.

1/3

• C.

1

• D.

0

A. 2/3
Explanation
If the probability of success is twice the probability of failure, it means that the probability of success is twice as likely to occur compared to the probability of failure. Let's assume the probability of failure is x. Since the probability of success is twice the probability of failure, it would be 2x. The total probability of all outcomes must equal 1. Therefore, x + 2x = 1. Solving for x, we get x = 1/3. So, the probability of success is 2/3.

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• 20.

### If Ø is an impossible event, then  P(Ø) =

• A.

0

• B.

1

• C.

1/4

• D.

1/2

A. 0
Explanation
If Ø is an impossible event, it means that it cannot occur. In probability theory, the probability of an impossible event is always 0. Therefore, P(Ø) = 0.

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• Dec 03, 2013
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