Independent Events Worksheets

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Independent Events Worksheets

ependent Events Worksheets

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1.

A box consists cards labeled 1 through 10. A card is drawn at random and replaced.

Then a second card is drawn. Find the probability that the first number is odd and the

second number is a multiple of 3.

a. 0.08

b. 1.5

c. 0.15

d. 0.8

Solution:

P (1st number is odd) = 510 = 1 / 2[There are 5 odd numbers: 1, 3, 5, 7, 9.]

P (2nd number is multiple of 3) = 310[There are 3 multiples of three: 3, 6, 9.]

P (1

st

number is odd and 2

nd

number is multiple of 3)

= P (1st number is odd) P (2nd number is multiple of 3)

= 12 310

= 320 = 0.15

= 320 = 0.15

The probability that the first number is odd and the second number is a multiple of 3 is 0.15.

2.

A box consists cards labeled 1 through 10 written on them. A card is drawn at

random and replaced. Then a second card is drawn. Find the probability that the first

number is less than 5 and second number is a prime number.


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a. 1.6

b. 0.08

c. 0.8

d. 0.16

Solution:

P (1st number less than 5) = 410[There are 4 numbers that are less than 5 : 1, 2, 3, 4]

P (2nd number is prime) = 410

[There are 4 prime numbers : 2, 3, 5, 7.]

P (1st number is less than 5 and 2nd number is prime) = P (1st number is less than 5) P(2ndnumber is prime)

= 410 410

= 0.16

The probability that the first number is less than 5 and second number is a prime number is

0.16.

3.

The probability that A gets a fellowship is 0.3 and B gets fellowship is 0.8. Find the

probability that both A and B get fellowship.

a. 0.24

b. 0.14

c. 0.11

d. 2.4


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Solution:

P(A gets fellowship) = 0.3

P(B gets fellowship) = 0.8

P(A and B get fellowship) = P(A gets fellowship) P(B gets fellowship)[A and B are independent events.]

= 0.3 0.8 = 0.24

4.

If P(A) = 0.1, P(B) = 0.4 and P(AB) = 0.2, then find P(AB)

a. 0.3

b. 0.7

c. 0.5

d. 0.6

Solution:

P (AB) = 0.1 + 0.4 0.2[P (AB) = P (A) + P (B) - P (AB).]

= 0.3

5.

If S is a sample space and A, B and C are mutually exclusive, then P(ABC) = ?

a. 1

b. P (A) + P (B) + P(C) - P(ABC)

c. P (A) + P (B) + P(C)


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Solution:

P (ABC) = P (A) + P (B) + P(C) P (AB) - P (BC) - P(CA) + P (ABC)

= P (A) + P (B) + P(C).[ A, B and C are mutually exclusive.

P (AB) = 0; P (BC) = 0; P (CA) = 0 and P (ABC) = 0.]

6.

If P (A) = 0.6, then P (A) = ?

a. 0.4

b. 0.6

c. 1

Solution:

P (A) = 1 - P (A)

= 1 - 0.6

= 0.4

7.

If P (E) = 0.02, then P (E) = ?

a. 0.08

b. 0.98

c. 0.02

d. 0.8


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Solution:

P (E) = 1 - P (E)

= 1 - 0.02

= 0.98

8.

If P(A) = 0.3, P(AB) = 0.06 and the events A and B are independent, then find P(B)?

a. 0.2

b. 0.02

c. 0.24

d. 0.03

Solution:

P(AB) = P(A) P(B)[A and B are independent events.]

0.06 = 0.3 P(B)

0.2 = P(B)

9.

If P (A) = 0.25 and P (B) = 0.5, then find P(AB), if A and B are independent events.

a. 1.25

b. 0.3

c. 0.75

d. 0.125


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Solution:

P(AB) = 0.25 x 0.5[P(AB) = P(A) x P(B).]

= 0.125

10.

A box consists cards labeled numbers 1 through 10. A card is drawn at random and

replaced. Then a second card is drawn. Find the probability that the first number

drawn is 5 and the second number is 10.

a. 0.1

b. 0.01

c. 1

d. 0.2

Solution:

Let A be the event of getting a number 5.

P(A) = 110

Let B be the event of getting a number 10.

P(B) = 110

P(the first number is 5 and the second number is 10) = P(A B)

= P(A) P(B)[A and B are independent events.]

= 110 110

= 1100 = 0.01

The probability that the first number drawn is 5 and the second number is 10 is 0.01.

11.


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A die is rolled and a coin is tossed. Find the probability of getting an odd number and

heads.

a. 112

b. 14

c. 12

d. 13

Solution:

Let A be the event of getting an odd number.

P(A) = 36 = 12[There are three odd numbers: 1, 3, and 5.]

Let B be the event of getting a head

P(B) = 12

P(AB) = P(A) P(B)[A and B are independent events.]

= (12)(12)

= 14

12.

Find the probability that a card drawn from a bridge deck may be either an ace or a

king.

a. 213

b. 113

c. 1169


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d. 126

Solution:

The number of ways of drawing a card from bridge deck = 52C1

The number of outcomes in sample space =52C1 = 52

Let A be the event of drawing an ace card

The number of outcomes in A = 4C1 = 4

Therefore, P (A) = 452 = 113

[P (E) = Number of outcomes in the event / Number of outcomes in the sample space.]

Let B be the event of drawing a king card.

The number of outcomes in B = 4C1 = 4

Therefore, P (B) = 452 = 113[P (E) = Number of outcomes in the event / Number of outcomes in the sample space.]

P (card drawn is either an ace or a king)

= P (AB)

= P (A) + P (B)[A and B are mutually exclusive events.]

= 113 + 113

= 213

The probability that a card drawn from a bridge deck may be either an ace or a king = 213.

13.

When two dice are rolled, what is the probability that the sum of dots on them is 8 or

9?


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a. 13

b. 14

c. 1518

d. 16

Solution:

The number of outcomes in sample space = 6 6 = 36

Let A be the event of getting the sum as 8.

A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

The number of outcomes in A = 5.

Therefore, P(A) = 536[P (E) = Number of outcomes in the event / Number of outcomes in the sample space.]

Let B be the event of getting the sum as 9

B = {(3, 6), (4, 5), (5, 4), (6, 3)}

The number of out comes in B = 4.

Therefore, P (B) = 436[P (E) = Number of outcomes in the event / Number of outcomes in the sample space.]

P (Sum of dots on two dice is 8 or 9)

= P (AB)

= P (A) + P (B)[A and B are mutually exclusive events.]

= 536 + 436 = 936 = 14.

When two dice are rolled, the probability that the sum of dots on them is 8 or 9 = 14.


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14.

A box consists of 5 white and 4 black balls. A ball is selected at random and is not

replaced. A second ball is selected. Find the probability that the first one is white and

the second one is black.

a. 1

b. 611

c. 518

d. 13

Solution:

P(selecting a white ball first) = 59[Total balls are 9.]

P(selecting a black ball next) = 48 = 12[Total balls are 8.]

P(selecting a white ball first and a black ball next)

= 59 12 = 518

The probability that the first one is white and the second one is black is 518.

15.


replaced. A second ball is selected. Find the probability that both the balls are white.

a. 518

b. 13

c. 611


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d. 1

Solution:

P (selecting a white ball first) = 59[Total balls are 9.]

P (selecting a white ball next) = 48 = 1 / 2[Total balls are 8.]

P (both the balls are white) = 59 12

= 518

The probability that both the balls are white is518.

16.


replaced. A second ball is selected. Find the probability that both the balls are black.

a. 1681

b. 717

c. 16

d. 5972

Solution:

P (selecting a black ball first) = 49

[Total balls are 9.]

P (selecting a black ball next) = 38[Total balls are 8.]

P (both the balls are black) = 49 38 = 16


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The probability that both the balls are black is16.

17.

If P (AB) = 34, P (A) = 23, P (B) = 23, then findP (AB).

a. 712

b. 14

c. 29

d. 13

Solution:

P (A) = 1 - P (A) = 1 - 23 = 13.

P (AB) = P (A) + P (B) P (AB)

34 = 13+ 23 P (AB)

34 = 33- P (AB)

P (AB) = 1 34 = 14

18.

Two cards are drawn at random from a 52-card deck. What is the probability that both

are red or both are jacks?

a. 4C252C2

b. 26C252C2 + 4C252C2 - 152C2

c. 26C252C2 + 4C252C2

d. 26C252C2


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Solution:

The number of ways of drawing 2 cards from a 52 card deck = 52C2

The number of outcomes in sample space =52C2

Let A be the event that both cards drawn are red

The number of outcomes in A = 26C2 ; P (A) =26C252C2.

Let B be the event that both cards drawn are Jacks

The number of outcomes in B = 4C2 ; P (B) =4C252C2

AB is the event that both the Jacks drawn are red

The number of outcomes in AB = 2C2 = 1; P (AB) = 152C2

P (Two cards drawn either red or jacks) = P(AB)

= 26C252C2 + 4C252C2 - 152C2[P (A) + P (B) - P (AB).]

19.

A bag contains tickets labeled 1 through 10. A ticket is selected at random and is not

placed before a second ticket is selected. Find the probability that the sum of the

numbers is less than 6.

a. 120

b. 225

c. 445

d. 12


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Solution:

The number of outcomes in the sample space = 10 9 = 90.

Let E be the event that the sum of numbers on the tickets is less than 6.

E = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (3, 1), (3, 2), (4, 1)}

The number of outcomes in E = 8.

P(E) =Number of outcomes in the eventNumber of outcomes in the sample space

= 890= 445

The probability that the sum of the numbers is less than 6 is 445.

20.

A bag contains tickets numbered 1 to 10. A ticket is selected at random and is

replaced before a second ticket is selected. Find the probability that the sum of the

numbers is greater than 15.

a. 320

b. 325

c. 34

d. 110

Solution:

The number of outcomes in sample space = 10 10 = 100

Let E be the event that the sum of numbers on the tickets is greater than 15.

E = {(6, 10), (7, 9), (7, 10), (8, 8), (8, 9), (8, 10 ), (9, 7), (9, 8), (9, 9), (9, 10), (10, 6), (10, 7),(10, 8), (10, 9), (10, 10)}.

The number of outcomes in E = 15.



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= 15100= 320

The probability that the sum of the numbers is greater than 15 is 320.

1.

A card is drawn at random from a bridge deck. What is the probability that the card is

either aface card (Jack, Queen, King) or an ace?

a. 113

b. 213

c. 313

d. 413

Solution:

The number of outcomes in sample space =52C1 = 52

Let A be the event that the card drawn is a face card

The number of outcomes in A = 12C1 = 12


P(A) = 1252= 313

Let B be the event that the card drawn is an ace

The number of outcomes in B = 4C1 = 4


P(B) = 452= 113

P(the card drawn is either a face card or an ace)

= P(AB)


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= P(A) + P(B)[A and B are mutually exclusive events.]

= 313+ 113

= 413

The probability that the card is either a face card (Jack, Queen, King) or an ace = 413.

22.

100 students appeared for two tests. 70 passed in the first, 50 passed in the second

and 40 passed in both. Find the probability that a student selected at random passed

in at least one test.

a. 15

b. 720

c. 45

d. 35

Solution:

Let A be the event that the students passed in the first test

The number of outcomes in A = 70

Therefore, P(A) = 70100

Let B be the event that the students passed in the second test.

The number of outcomes in B = 50

Therefore, P (B) = 50100

AB is the event that the student has passed both the tests.

The number of outcomes in AB = 40


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Therefore, P (AB) = 40100

P (student passed in at least one test)

= P (AB) = 70100 + 50100 40100[P(AB) = P(A) + P(B) - P(AB).]

= 80100

= 45

The probability that a student selected at random passed in at least one test = 45.

23.

If P (A) = 35; P (B) = 25; P(C) = 35;P (AB) = 310; P (BC) = 15;P(CA) = 25; P

(ABC) = 15, then find P (ABC).

a. 35

b. 45

c. 910

d. 310

Solution:

P (ABC) = P (A) + P (B) + P(C) - P (AB) - P (BC) - P(CA) + P (ABC)

= 35 + 25 + 35 - 310 - 15 - 25 + 15

= 65 310

= 910

24.


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A bag contains 3 red and 4 blue pens, another contains 2 red and 3 blue pens. One of

the bags is chosen at random and two pens are drawn from it. Find the probability

that the two pens are blue.

a. 4170

b. 310

c. 27

d. 41140

Solution:

Let E1 be the event of selecting the first bag and E2 be the event of selecting the secondbag.

P (E1) = 12, P (E2) = 12

Let A be the event that the two pens drawn are both blue.

P (A) = P (E1) P (A given E1) + P (E2) P (A given E2)

= 12 4C27C2 + 12 3C25C2

= 12 621 + 12 310

=17 + 320 = 41140

The probability that the two pens are blue =41140.

25.

A card is drawn from a bridge deck and is not replaced before a second card isdrawn. Find the probability that both cards drawn are face cards.

a. 352

b. 11221


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c. 313

d. 1442

Solution:

P (both cards drawn are face cards)

= (1252) x (1151)[There are 12 face cards in a 52-card deck.]

= 11221

26.

A bag contains balls labeled 1 through 20. One ball is chosen at random from the bag

and its number is noted. Let A be the event that the number is divisible by 2, B be the

event that the number is divisible by 3, and C be the event that the number is greater

than 5 and less than 15. Find P (ABC).

a. 19200

b. 54

c. 45

d. 710

Solution:

P (A) = 1020 = 12

[There are 10 numbers divisible by 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.]

P (B) = 620 = 310[There are 6 numbers divisible by 3: 3, 6, 9, 12, 15, and 18.]

P(C) = 920[There are 9 numbers between 5 and 15.]


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P (AB) = 3 / 20[There are 3 numbers which are divisible by 2 and 3: 6, 12, 18.]

P (BC) = 3 / 20[There are 3 numbers which are divisible by 3, greater than 5 and less then 15: 6, 9, 12.]

P(CA) = 520 = 14[There are 5 numbers which are divisible by 2 and lie between 5 and 15: 6, 8, 10, 12, and14.]

P (ABC) = 220 = 110[There are 2 numbers which are divisible by both 2 and 3 and that lie between 5 and 15: 6,12]

P (ABC) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(CA) + P(ABC)

= 1620 = 45[Substitute and simplify.]

P (ABC) = 1620 = 45.

27.

Two events A and B are mutually exclusive andP(A) = 14, P(B) = 12. Find P(AB).

a. 13

b. 18

c. 58

d. 34

Solution:

P (AB) = 14 + 12[P (AB) = P (A) + P (B) when A and B are mutually exclusive.]

= 34


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Independent Events Worksheets

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