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**Short Answer Type Questions**

**Q.1. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5. The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains II green marbles, find the total number of marbles in the jar. [CBSE, Allahabad 2019]Ans. ** Let the total number of marbles in the jar = x

Probability of selecting a blue marble + Probability of selecting a black marble + Probability of selecting a green marble = 1

(i) is a prime number

(ii) lies between 2 and 6. [Delhi 2019]

Ans.

Prime numbers are 2, 3 and 5.

(i) Probability of getting a prime number = 3/6 = 1/2.

(ii) Probability of getting a number lies between 2 and 6.

= 3/6 = 1/2.

A B C A A B

(i) Let E be the event of getting a letter A .

∴ Favourable number of elementary events = 3

∴

(ii) Let E be the event of getting a letter B.

∴ Favourable number of elementary events = 2

∴

Ans.

{(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)}

Total elementary events = 36

(i) Outcomes of doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, i.e. 6

P(getting a doublet) = 6/36 = 1/6

(ii) Outcomes of getting a sum 10 = {(4, 6), (5, 5), (6, 4)}, i.e. 3

P (getting a sum 10) = 3/36 =1/12.

Find the probability that it is:

(i) divisible by 8.

(ii) not divisible by 8. [CBSE 2018]

Ans.

Total outcomes = 98

P(divisible by 8) =

(ii) P (not divisible by 8)

Ans.

∴ Total number of elementary events = 900

One rotten apple is randomly selected from this heap

⇒ Favourable events = 0.18 x 900 = 162

Hence, there are 162 rotten apples in the heap.

Ans.

Possible outcomes

{HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}

Number of 2 heads together = 3

Probability of getting two head = 3/8

∴ Total number of possible outcomes = 8

Now, Ramesh will lose the game if he gets HHT, HTH, THH, TTH, THT, HTT.

∴ Favourable number of events = 6

∴ Probability that he lose the game = 6/8 = 3/4

(i) have a sum less than 7

(ii) have a product less than 16

(iii) is a doublet of odd number. [Delhi 2017]

Ans.

(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)

Total number of elementary events = 6 x 6 = 36

(i) Let A be the event of getting a sum less than 7 in the numbers obtained on the two dice.

∴ Elementary events favourable to event A are:

(1,1), (2,1), (3,1), (4,1), (5,1), (1,2), (1,3), (1,4), (1,5), (2,2), (2,3), (2,4), (3,2), (3,3), (4,2)

Favourable number of elementary events = 15

Hence, required probability = 15/36 = 5/12

(ii) Let A be the event of getting a product less than 16

∴ Elementary events favourable to event A are

(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), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2)

⇒ Favourable number of elementary events = 25

Hence, required probability = 25/36

(iii) Let A be the event of doublet of odd numbers. Elementary events favourable to the event A are (1,1), (3,3), (5,5)

∴ Favourable number of elementary events = 3

∴ Required probability = 3/36 = 1/12.

Ans.

Total outcomes = 7

∴ Required probability = 3/7

Ans.

∴ P (perfect square number) = 6/48 or 1/8

Ans.

Number of red cards and queens = 28

Number of favourable outcomes = 52 - 28 = 24

P (getting neither a red card nor a queen) = 24/52 = 6/13

Ans.

∴ Required probability = 8/20 or 2/5

(i) all heads.

(ii) exactly two heads.

(iii) exactly one head.

(iv) at least two heads.

(v) at least two tails.

Ans.

∴ Total number of elementary events = 8

(i) The event “getting all heads” is said to occur, if the elementary event HHH occurs, i.e. favourable number of elementary events = 1

Hence, required probability = 1/8

(ii) The event “getting two heads” will occur, if one of the elementary events HHT, THH, HTH occurs i.e.; favourable number of elementary events = 3

Hence, required probability =3/8

(iii) The event of “getting one head”, when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs i.e. favourable number of elementary events = 3

Hence, required probability = 3/8

(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that the event “getting at least two heads” occurs i.e.; favourable number of elementary events = 4

Hence, required probability = 4/8 = 1/2.

(v) Similar as

(iv) P (getting at least two tails) = 4/8 = 1/2

Ans.

(i) Favourable outcomes are (2,2) (2,3) (2,5) (3,2) (3,3) (3,5) (5,2) (5,3) (5,5) i.e., 9.

P (a prime number on each die) = 9/36 or 1/4

(ii) Favourable outcomes are (3,6) (4,5) (5,4) (6,3) (5,6) (6,5) i.e., 6 outcomes

P (a total of 9 or 11) = 6/36 or 1/6

**Long Answer Type Questions**

**Q.1. A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that number on the drawn card is(i) a prime number(ii) a composite number(iii) a number divisible by 3**(i) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19

Number of prime numbers (favourable cases) = 8

Total possible outcomes = 20

P(prime number) = 8/20 = 2/5

(ii) Composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20

Number of composite numbers (Favourable outcomes) = 11

Total possible outcomes = 20

P(composite numbers) =11/20

(iii) Numbers divisible by 3 are 3,6,9,12,15 and 18

Number of favourable outcomes = 6

Total possible outcomes = 20

P(number divisible by 3) = 6/30 = 3/10

(i) of spade

Ans.

Remaining cards = 52 - 3 = 49

(i) Number spade = 13

P(spade) = 13/49

(ii) Number of black king = 1

P (black king) =1/49

(iii) Number of clubs = 10

P(clubs) = 10/49

(iv) Number of Jacks = 3

P(Jacks) =3/49

(i) two digit number.

(ii) perfect square number. [CBSE 2017(C)]

Ans.

Two digit numbers are from 10 to 90

Number of two digit cards = 81

(i) Probability of getting 2-digit number = 81/90 = 9/10

(ii) Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64,81.

∴ number of perfect square = 9

Probability of getting perfect square = 9/90 = 1/10

(i) even sum

(ii) even product. [CBSE (AI) 2017]

Ans.

Number of outcomes when sum is even = 18 [(1, 1), (1, 3) ...(6, 6)]

Number of outcomes when product is even = 27 [(1, 2), (1, 4) ... (6, 6)]

(i) P (even sum) =18/36 = 1/2

(ii) P (even product) =27/36 = 3/4

Rina throws only one die Both Peter and

Rina want the same outcome 25 but their approach is not the same.

Peter will multiply the outcomes on the two dice whereas Rina will square the number obtained on throwing only one die.

let us calculate the probability in each case when a pair of die is thrown there are 36 elementary events as given below:

(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)

The product of two numbers on the two dice will be 25 if both the dice show number 5.

So, there is only one elementary event favourable to to get a ‘5’ on each die.

∴ P(Peter getting 25) =1/36

Rina throws only one die on which she can get one of the six numbers 1, 2, 3, 4, 5, 6.

If she gets number 5 on the upper face of the die thrown, then the square of the number is 25.

∴ P

Clearly P

So, Rina has better chance of getting the number 25.

Total number of cases of product of x and y = 16

Product less than 16 = (1 x 1, 1 x 4, 1 x 9, 2 x 1, 2 x 4, 3 x 1, 3 x 4, 4 x 1)

Number of cases, where product is less than 16 = 8

∴ Required probability = 8/12 or 1/2

Ans.

a = 2, b can take 1 value,

a = 3, b can take 2 values,

a = 4, b can take 3 values,

a = 5, b can take 4 values,

a = 6, b can take 5 values.

Total possible outcomes = 36

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