Probability Exercise 13.1(Part-3) | Extra Documents, Videos & Tests for Class 10 PDF Download

Q: 43. A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be

(i) a blue card

(ii) not a yellow card

(iii) Neither yellow nor a blue card.

 Solution:

Total no. of possible outcomes = 100 + 200 + 50 = 350 {100 red, 200 yell= & SO blue}

(i)  E = event of getting blue card.

No. of favorable outcomes = 50 (50 blue cards}

P(E) = 50/350

P(E) = 1/7

(ii)  E = event of getting yellow card

No. of favorable outcomes = 200 (200 yellow)

P(E) = 200/350

P(E) = 4/7

event of not getting yellow card 

= 3/7

(iii)  E =  getting neither yellow nor a blue card

No. of favorable outcomes = 350 – 200 – 50 = 100 {removing 200 yellow & 50 blue cards) P(E) =100/350

P(E) = 2/7

Q: 44. The faces of a red cube and a yellow cube are numbered from 1 to 6. Both cubes are rolled. What is the probability that the top face of each cube will have the same number.

Solution:    Total no. of outcomes when both cubes are rolled = 6 x 6 = 36 which 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) (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)}

E = event of getting same no. on each cube

No. of favorable outcomes = 6 which are

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

probability, P(E) = 

P(E) = 6/36

= 1/6

Q: 45. The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is 1/4.  The probability of selecting a white marble at random from the same jar is 1/3. If this jar contains 10 yellow marbles. What is the total number of marbles in the jar?

Sol:

Let the no. of green marbles = x

The no. of white marbles = y

No. of yellow marbles =10

Total no. of possible outcomes = x+ y+ 10 (total no. of marbles)

P(E)= 

Probability (green marble) = 

x + y+ 10 = 4*x

3 * x – y – 10 = 0 …….. (i) 

Probability (White marble) = 

x + y+ 10 = 3 * y

x +y+10 = 3 * y

x - 2 * y +10 = 0…………. (ii) 

Multiplying (ii) by 3,

3 * x - 6 * y  + 30 = 0 ……… (iii)

Sub (i) from (iii), we get 

5 * y = 40

y= 8

Substitute y=8 in (i),

Total number of marbles in jar= x +y + 10 = 6+ 8 + 10=24 

Q:46.  (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (a) is not defective and not replaced. Now bulb is drawn at random from the rest. What Is the probability that this bulb is not defective?

Solution:

Total no. of possible outcomes = 20 {20 bulbs}

(i)  E = event of getting defective bulb.

No. of favorable outcomes = 4 (4 defective bulbs)

Probability, P(E) = 

P(E) = 4/20

P(E) = 1/5

(ii)  Bulb drawn in is not detective & is not replaced

Remaining bulbs = 15 good + 4 bad bulbs = 19

Total no. of possible outcomes =19

E = event of getting defective

No. of favorable outcomes = 15 (15 good bulbs)

P(E) =15/19

Q: 47. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears:

(i) a two digit number

(ii) a perfect square number

(iii) a number divisible by 5.

Solution:

Total no. of possible outcomes -90 {1, 2, 3 ……. 90}

(i) E = event of getting 2 digit no.

No. of favorable outcomes = 81 {10, 11, 12, …. 90}

Probability, P(E) = 

P(E) = 81/90

P(E) = 9/10

(ii) E = event of getting a perfect square.

No. of favorable outcomes = 9 {1, 4, 9, 16, 25, 26, 49, 64, 81}

P(E) = 9/90

P(E) = 1/10

(iii) E = event of getting a no. divisible by 5.

No. of favorable outcomes = 18 {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}

P(E) =18/90

P(E) = 1/5

Q: 48. Two dice, one blue and one grey, are thrown at the same time. Complete the following table: 

From the above table a student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7,  8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? 

Solution:

Total no. of possible outcomes when 2 dice are thrown = 6 x 6 = 36 which 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) (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)}

E = event of getting sum on 2 dice as 2

No. of favorable outcomes = 1{(1, 1)}

Probability, P(E) =  

P(E) =  1/36

E = event of getting sum as 3

No. of favorable outcomes = 2 ((1, 2) (2, 1))

P(E) = 2/36

P(E) = 1/18

 E = event of getting sum as 4

No. of favorable outcomes = 3 {(3, 1) (2, 2) (1, 3)}

P(E) =3/36

P(E) =  1/12

 E = event of getting sum as 5 

No. of favorable outcomes = 4 {(1,  4) (2, 3) (3, 2) (4, 1)} 

P(E) =4/36

P(E) = 1/9

 E = event of getting sum as 6

No. of favorable outcomes = 5 {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}

P(E) =5/36

E = event of getting sum as 7

No. of favorable outcomes = 6 {(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)}

P(E) =6/36

P(E) = 1/6

 E = event of getting sum as 8

No. of favorable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

P(E) =5/36

E = event of getting sum as 9

No. of favorable outcomes = 4 {(3, 6) (4, 5) (5, 4) (6, 3)}

P(E) =4/36

P(E) = 1/9

E = event of getting sum as 10

No. of favorable outcomes = 3 {(4, 6) (5, 5) (6, 4)}

P(E) =3/36

P(E) = 1/12

Q: 49. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag.

Solution:

No of red balls = 6

Let, no. of blue balls = x

Total no. of possible outcomes = 6+ x (total no. of balls)

P(E) =

P(Blue ball) = 2P(red ball) 

x = 12

Therefore, number of blue balls = 12

Q: 50. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of  

(i) heart

(ii) queen

 (iii) clubs. 

Solution:

Total no. of remaining cords = 52 – 3 = 49

(i) E = event of getting hearts

No. of favorable outcomes = 13 {13 hearts}

Probability, P(E) = 

P(E) = 13/49

(ii)  E= event of getting queen

No. of favorable outcomes = 3 {4 – 1} (Since queen of clubs is removed)

P(E) =3/49

(iii)  E = event of getting clubs

No. of favorable outcomes =10 (13 – 3) {Since 3 club cards are removed)

P(E) =10/49

E = event of getting sum as 11

No. of favorable outcomes = 2 {(5, 6) (6, 5)}

P(E) =2/36

P(E) = 1/18

E = event of getting sum as 12

No. of favorable outcomes =1 {(6, 6)}

P(E) = 1/36

No, the outcomes are not equally likely. From the above results we see that, there is different probability for different outcome. 

Q: 51. Two dice are thrown simultaneously. What is the probability that:

(i) 5 will not come up on either of them?

(ii) 5 will come up on at least one?

(iii) 5 will come up at both dice?

Solution:

Total no. of possible outcomes when 2 dice are thrown = 6 x 6 = 36 Which 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) (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)}

(i) E = event of 5 not coming up on either of them

No. of favorable outcomes = 25 Which 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) (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)}

Probability, P(E) = 

P(E) = 25/36

(ii) E = event of 5 coming up at least once

Number of favorable outcomes = 11 Which are

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

P(E) = 

P(E) = 11/36

(iii)  E = event of getting 5 on both dice

No. of favorable outcomes =1 {(5, 5)}

P(E) = 1/36

Q: 52. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.

Solution:

Total no. of possible outcome = 50 {1, 2, 3 …. 50}

No. of favorable outcomes = 4 {12, 24, 36, 48}

P(E) = 

P(E) =4/50

P(E) = 2/25