Ex1. A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability (P) that the ball drawn is:

(i) a black ball.

(ii) a red ball.

(iii) a white ball.

(iv) not a red ball.

(v) not a black ball.

solutions:

Total number of balls =3+5+2=10

(i) There are 5 black balls

*P*(getting a black ball)=5/10=½

(ii) There are 2 red balls

*P*(getting a red ball)=2/10=⅕

(iii) There are 3 white balls

*P*(getting a white ball)=3/10

(iv) There are 3+5=8 balls which are not red

*P*(getting a non red ball)=8/10=⅘

(v) There are 3+2=5 balls which are not black

Favourable number of events =*P*(A)=5

*P*(getting a non black ball)=5/10=½

Ex2. A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:

(i) a white ball

(ii) a black ball

solutions:

Possible number of outcomes =6= number of balls in the bag, *n*(S)=6

(i) E= event of drawing a white ball =number of white balls in the bag =0

*n*(E)=0

Probability of drawing a white ball

*P*(E)=

*n*(E)/

*n*(S) =0/6=0

(ii) E= event of drawing a black ball =number of black balls in the bag =6

*n*(E)=6

Probability of drawing a black ball

*P*(E)=

*n*(E)/

*n*(S) =6/6=1

Ex3. A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is:

(i) red (ii) not red (iii) white (iv) not white (v) green or red (vi) white or green (vii) green or red or white

solutions:

Balls in the bag =16= Number of balls that could be drawn, *n*(S)=16.

(i) E= Event of drawing a red ball =number of red balls =7, *n*(E)=7.

Probability of drawing a red ball =*n*(E)/*n*(S) =7/16

(ii) Number of not red balls =*n*(E)=6+3=9.

Probability of drawing a not red ball =*n*(E)/*n*(S) =9/16

(iii) E= Event of drawing a white ball =number of white balls =*n*(E)=3

Probability of drawing a white ball =*n*(E)/*n*(S) =3/16

(iv) Number of not white balls =6+7=13

Probability of not drawing a white ball =*n*(E)/*n*(S) =13/16

(v) E= Event of drawing a green or a red ball =number of green balls+number of red balls =6+7=13

Probability of drawing a green or a red ball =*n*(E)/*n*(S) =13/16

(vi) E= Event of drawing a green or a white ball =number of green balls+number of white balls =6+3=9

Probability of drawing a green or a white ball =*n*(E)/*n*(S) =9/16

(vii) E= Event of drawing a green or a white or a red ball =number of green balls+number of white balls+number of red balls. *n*(E)=16.

Probability of drawing a green or a white or a red ball =*n*(E)/*n*(S) =16/16=1, that is the probability of each sure occurrence as each element in the sample space.

Ex4. A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is:

(i) white (ii) red (iii) not green (iv) red or white

solutions:

Total number of balls in the box =48

Total possible outcomes on drawing a ball =48. *n*(S)=48.

(i) Event of drawing a white ball, *n*(E)=12

Probability of drawing a white ball =*n*(E)/*n*(S) =12/48=¼

(ii) Event of drawing a red ball, *n*(E)=16

Probability of drawing a red ball =*n*(E)/*n*(S) =16/48=⅓

(iii) Event of drawing a green ball, *n*(E)=20

Probability of drawing a green ball =*n*(E)/*n*(S) =20/48=5/12

Probability of not drawing a green ball =1-5/12=7/12

(iv) red or a white ball =12-16=28 balls

Event of drawing a red or white ball, *n*(E)=28

Probability of not drawing a green ball =*n*(E)/*n*(S) =28/48=7/12

Ex5. A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:

(i) yellow

(ii) red

(iii) not yellow

(iv) neither yellow nor red

solutions:

Total number of balls in the bag =3+4+1=8 balls

Number of possible outcomes =8=*n*(S)

(i) Event of drawing a yellow ball *n*(E)=1

Probability of drawing a yellow ball =*n*(E)/*n*(S) =⅛

(ii) Event of drawing a red ball *n*(E)=3

Probability of drawing a red ball =*n*(E)/*n*(S) =⅜

(iii) Probability of not drawing a yellow ball =1- Probability of drawing a yellow ball

Probability of not drawing a yellow ball =1-⅛=(8-1)/8=⅞

(iv) Neither yellow ball nor red ball means a blue ball

Event of not drawing a yellow or red ball *n*(E)=4

Probability of not drawing a yellow or red ball

*P*=*n*(E)/*n*(S) =4/8=½

Ex6. A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:

(i) white

(ii) neither red nor white. Solution:

solutions:

Total number of balls in the box =7+8+5=20 balls Total possible outcomes *n*(S)=20

(i) Event of drawing a white ball =E= number of white balls *n*(E)=5

Probability of drawing a white ball =*n*(E)/*n*(S) =5/20=¼

(ii) Neither red ball nor white ball =green ball

Event of not drawing a red or white ball =E= number of green ball *n*(E)=8

Probability of drawing a white ball =*n*(E)/*n*(S) =8/20=⅖

Ex7. A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:

(i) black (ii) red

solutions:

Total possible outcomes =number of red balls.

(i) Number of favorable outcomes for black balls =0

*P*(black ball)=0

(ii) Number of favorable outcomes for red balls =number of red balls, *n*(E)=*n*(S)

*P*(red ball)=1

Ex8. A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:

(i) not red?

(ii) neither red nor green?

(iii) white or green?

solutions:

Total number of possible outcomes =10+16+8=34 balls *n*(S)=34

(i) Favorable outcomes for not a red ball =favorable outcomes for white or green ball

Number of favorable outcomes for white or green ball =16+8=24=*n*(E)

Probability for not drawing a red ball =*n*(E)/*n*(S) =24/34=12/17

(ii) Favorable outcomes for neither a red nor a green ball =favorable outcomes for white ball

Number of favorable outcomes for white ball =16=*n*(E)

Probability for not drawing a red or green ball

*P*=

*n*(E)/

*n*(S) =16/34=8/17

(iii) Number of favorable outcomes for white or green ball =16+8=24=

*n*(E)

Probability for drawing a white or green ball

*P*=

*n*(E)/

*n*(S) =24/34=12/17

Ex9. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

solution:

Total number of possible outcomes *n*(S)=18 {10 red balls, 8 white balls}

E→ event of drawing white ball

Number of favorable outcomes *n*(E)=8 {8 white balls}

*P*=

*n*(E)/

*n*(S) =8/18=4/9

Ex10. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) White (ii) Red (iii) Black (iv) Not red

Solutions:

Total number of possible outcomes *n*(S)=12 {3 red balls, 5 black balls & 4 white balls}

(i) E→ event of getting white ball

Number of favorable outcomes *n*(E)=4 {4 white balls}

Probability, *P*=4/12=⅓

(ii) E→ event of getting red ball

Number of favorable outcomes *n*(E)=3 {3 red balls}

*P*=

*n*(E)/

*n*(S) =3/12=¼

(iii) E→ event of getting black ball

Number of favorable outcomes =5 {5 black balls}

*P*(E)=5/12

(iv) E→ event of getting red

Number of favorable outcomes =3 {3 black balls)

*P*(E)=3/12=¼

Ē→ event of getting a non-red.

*P*(Ē)=1-

*P*(E)

*P*(Ē)=1-¼=¾

Ex11. A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?

solution:

Total number of possible outcomes =12 {5 white, 7 red}

E→ event of drawing white ball.

Number of favorable outcomes *n*(S)=5 {white balls are 5}

*P*(E)=

*n*(E)/

*n*(S) =5/12

Ex12. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) Red (ii) black or white (iii) not black

Solutions:

Total number of possible outcomes *n*(S)=5+7+3=15 {5 black, 7 red & 3 white balls}

(i) E→ event of drawing red ball

Number of favorable outcomes *n*(E)=7 {7 red balls)

*P*(E)=

*n*(E)/

*n*(S) =7/15

(ii) E→ event of drawing black or white

Number of favorable outcomes

*n*(E)=8 (5 black & 3 white)

*P*(E)=

*n*(E)/

*n*(S) =8/15

(iii) E→ event of drawing black ball

Number of favorable outcomes

*n*(E)=5 {5 black balls)

*P*(E)=

*n*(E)/

*n*(S) =5/15=⅓

Ē→ event of not drawing black ball

*P*(Ē)=1-

*P*(E)

*P*(Ē)=1-⅓=⅔

Ex13. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) White (ii) Red (iii) Not black (iv) Red or white

Solutions:

Total number of possible outcomes *n*(S)=15 {4 red, 5 black, 6 white balls}

(i) E→ event of drawing white ball.

Number of favorable outcomes *n*(E)=6 {6 white)

*P*(E)=

*n*(E)/

*n*(S) =6/15=⅖

(ii) E→ event of drawing red ball

Number of favorable outcomes =4 {4 red balls)

*P*(E)=4/15

(iii) E→ event of drawing black ball

Number of favorable outcomes =5 {5 black balls)

*P*(E)=5/15=⅓

Ē→ event of not drawing black ball

*P*(Ē)=1-

*P*(E)

*P*(Ē)=1-⅓=⅔

(iv) E→ event of drawing red or white ball

Number of favorable outcomes =10 {4 red & 6 white)

*P*(E)=10/15=⅔

Ex14. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) Red (ii) Black

Solutions:

Total number of possible outcomes *n*(S)=8 (3 red, 5 black)

(i) Let E→ event of drawing red ball. *n*(E)=3

*P*(E)=

*n*(E)/

*n*(S) =⅛

(ii) Let E→ event of drawing black ball.

*n*(E)=5

*P*(E)=

*n*(E)/

*n*(S) =⅝

Ex15. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

(i) red or white (ii) not black (iii) neither white nor black.

Solutions:

Total number of possible outcomes *n*(S)=5+8+7=20 (5 red, 8 white & 7 black)

(i) E→ event of drawing red or white ball

Number of favorable outcomes *n*(E)=13 {5 red, 8 white)

*P*(E)=

*n*(E)/

*n*(S) =13/20=0.65

(ii) getting a non-black ball means getting a red ball or a white one.

*P*=13/20=0.65

(iii) Let E→ be event of getting neither white nor black ball. That means getting a red ball,

*n*(E)=5

*P*(E)=

*n*(E)/

*n*(S) =5/20=¼

Ex16. A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the hail.

Find the probability that the drawn ball is

(i) Red or white (ii) Not black (iii) Neither white nor black

Solutions:

Total number of possible outcomes *n*(S)=8+6+4=18 (8 red, 6 white, 4 black)

(i) Probability of getting a red or a white ball

*P*=(8+6)/18=7/9

(ii) getting a not black ball means getting a red or a white ball. The probability is

*P*=(8+6)/18=7/9

(iii) getting neither white nor black means getting a red ball. The probability is

*P*=8/18=4/9

Ex17. There are 5 green and 3 red balls in a box, one ball is taken out. find the probability in each case:

Events Happening:

(i). The ball is green

Solution:

Total number of balls =*n*(S)=5+3=8

One ball is taken as random.

Let event A denote that the ball is green. So, total possible outcomes = 8. Favorable outcomes =*n*(A)=5.

*P*(A)=

*n*(A)/

*n*(S) =⅝

(ii) The ball is red

Solution:

Total number of balls =

*n*(S)=5+3=8. One ball is taken as random. Let event B denote that the ball is red. So, total possible. Outcomes = 8. Favorable outcomes =

*n*(A)=3

*P*(B)=

*n*(B)/

*n*(S) =⅜

Ex18. A bag contains 40 balls out of 5 are green, 15 are black and the remaining are yellow. A ball is taken out of the bag. Find the probability in each case:

Event Happening:

(i). The ball is black

Solution:

Total balls *n*(S)=40

Green balls =5

Black balls =15

Yellow balls =40-20=20

Let event A denote that the ball is black.

Favourable outcomes =*n*(A)=15

*P*(A)=

*n*(A)/

*n*(S) =15/40=⅜

(ii). The ball is green

Solution:

Let event B denote that the ball is green.

Favourable outcomes =

*n*(B)=5

*P*(B)=

*n*(B)/

*n*(S) =5/40=⅛

(iii). The ball is not green

Solution:

Let event C denote that the ball is not green.

*P*(C)=1-

*P*(ball is green)=1-

*P*(B)

=1-⅛=⅞

## Algebraic inserts

Ex19. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

solution:

Let the number of blue balls be *x*.

Number of red balls =5

Total number of balls =*x*+5

Hence, number of blue balls =10

Ex20. A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.

solutions:

Let the number of black balls in the box be x. Total number of balls in the box =*x*+30

But,

*P*(drawing a black ball)=⅖×

*P*(drawing a white ball)

Number of black balls in the box =12

Ex21. A box contains 12 balls out of which *x* are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find *x*.

solution:

Total number of balls =12

Total number of black balls =*x*

*P*(getting a black ball)=*x*/12

If 6 more black balls are put in the box, then

Total number of balls =12+6=18

Total number of black balls =*x*+6

*P*(getting a black ball now)=(*x*+6)/18

According to the condition given in the question,

Total number of black balls =*x*

*P*(getting a black ball)=*x*/12

If 6 more black balls are put in the box, then

Total number of balls =12+6=18

Total number of black balls =*x*+6

*P*(getting a black ball now)=(*x*+6)/18

According to the condition given in the question,

Ex22. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball the bag is twice that of a red ball, find the number of blue balls in the bag.

solution:

Number of red balls =6

Let number of blue balls =*x*

Total number of possible outcomes =6+*x* (total number of balls)

Number of blue balls =12