Probability Test

Result

Probability Test - 6

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the drawn ticket contains a number, which is a multiple of 3 or 7?

A

B

C

D
None of these

Solution

Favourable cases = 3, 6, 9, 12, 15, 18, 7, 14
Probability =
Question 2
1 / -0

In a box, there are 5 red, 9 green and 6 black balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A

B

C

D

Solution

Total number of balls = 5 + 9 + 6 = 20
E = event that the ball drawn is neither red nor green = event that the ball drawn is black = 6
Therefore, P(E) =
Question 3
1 / -0

In a box, there are 5 Nestle and 18 Dairy Milk chocolates. Mohan picks up a chocolate without looking at it. What is the probability that it will be a Nestle chocolate?

A

B

C

D

Solution

Number of Nestle chocolates = 5
Number of Dairy Milk chocolates = 18
Total number of chocolates = 18 + 5 = 23
Probability that a picked up chocolate is Nestle =
Question 4
1 / -0

Two dice are rolled simultaneously. What is the probability that the total obtained is not a prime number?

A
5/12

B
7/12

C
11/18

D
1/2

Solution

S = Sample space
n(S) = 6 × 6 = 36
E = Event of getting a prime number = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
n(E) = 15
P(Not a prime number) = =
Question 5
1 / -0

A coin is tossed. If it shows heads, then a dice is rolled. All possible outcomes are equally likely. Find the probability that the dice shows a number greater than 5.

A

B

C

D

Solution

Probability that the coin shows heads =
S = {1, 2, 3, 4, 5, 6}
A = Event that dice shows a number greater than 5 = {6}
Required probability =
Question 6
1 / -0

A bag has 6 green marbles and 5 red marbles. Two marbles are drawn. What is the probability of getting both marbles of same colour, if two marbles are drawn without replacement?

A

B

C

D

Solution

Possible cases:
1. Green and Green
2. Red and Red
Probability of both green =
Probability of both red =
By addition theorem of probability,
Required probability = =
Question 7
1 / -0

Two fair dice are thrown. What is the probability that the number of dots on the first dice is greater than 3 and that on the second is greater than 4?

A

B

C

D

Solution

A: The first dice shows 4, 5 or 6 dots.
B: The seconds dice shows 5 or 6 dots.

P(A) =

P(B) =

So, the required probability is

P(A) × P(B) =
Question 8
1 / -0

A card is drawn from a pack of 52 cards. What is the probability of getting a non-face card or an ace card?

A
1/13

B
11/13

C
9/13

D
10/13

Solution

There are 40 non-face cards and 4 ace cards which are already counted in non face cards.
So, the total number of favourable cases are 40.
Required Probability = 40/52 = 10/13
Question 9
1 / -0

Suresh throws 2 dice simultaneously. What is the probability of getting two numbers whose product is odd?

A
5/36

B
1/6

C
1/4

D
1/2

Solution

In a two dice throw, the total number of cases are 36 i.e., n(S) = 36
Let E be the event of getting two numbers whose product is odd.
Then, E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
n(E) = 9
So, probability = 9/36 = 1/4
Question 10
1 / -0

From a deck of playing cards, one card is removed randomly & found to be a Jack of hearts. Then a second card is drawn without replacing the first one. What is the probability of getting a second card as a Jack of red colour?

A
1/52

B
1/51

C
11/52

D
13/51

Solution

The first card is a jack of red colour.
On the second turn, there is only one remaining Jack card of red colour (Jack of diamond) &
total no. of remaining cards is 51.
So, probability = 1/51
Question 11
1 / -0

A sample space consists of numbers from 0 to 20. What is the probability that an element of the sample space is an even number, excluding numbers 0 and 20?

A
9/20

B
3/7

C
11/21

D
10/21

Solution

Sample space is 21.
Even numbers between 0 and 20 are {2, 4, 6, 8, 10, 12, 14, 16, 18}
Number of favourable cases = 9
Hence, the probability is 9/21 = 3/7.
Question 12
1 / -0

A single card is drawn from a well-shuffled deck of playing cards. What is the probability of the card being black with an even number?

A
5/26

B
5/52

C
2/13

D
1/4

Solution

There are total five even numbers (2, 4, 6, 8 and 10). So, there are total 10 black cards (club and spade) with an even number on them.
Total number of cards = 52
So, required probability = 10/52 = 5/26
Question 13
1 / -0

Find the probability that a leap year selected at random contains either 53 Sundays or 53 Mondays.

A

B

C

D

Solution

A leap year has 366 days, or 52 weeks + 2 days. This means that 52 Sundays will be there, but for the 53^rd, we have to evaluate the possibility of the two extra or odd days to fall on Sunday or Monday.
The possible outcomes for two odd days can be (Mon, Tue); (Tue, Wed); (Wed, Thu); (Thu, Fri); (Fri, Sat); (Sat, Sun) and (Sun, Mon).
Number of favourable cases = 3
Required probability =
Question 14
1 / -0

In a dice game, a player rolls two dice. His score is the larger of the two numbers on the dice. For example, if he rolls 4 and 6, his score is 6 and if he rolls 3 and 3, his score is 3. What is the probability that his score is 4 or more?

A

B

C

D

Solution

When a player rolls two dice, there are 6 possibilities for the outcome on each die.
So, there are 36 possibilities for the outcomes when two dice are rolled.
The possibilities that give a score of 3 or less are 1 and 1, 1 and 2, 1 and 3, 2 and 1, 2 and 2, 2 and 3, 3 and 1, 3 and 2, 3 and 3.
So, 9 of the 36 possibilities give a score of 3 or less. And the probability of getting 3 or less =
Thus, probability of getting 4 or more = 1 -
Question 15
1 / -0

Three boxes A, B and C contain 4 red and 5 blue balls, 3 red and 6 blue balls and 7 red and 4 blue balls, respectively. If a ball drawn at random is blue, then find the probability that it was drawn from box C.

A

B

C

D

Solution

P (A) = Probability of choosing box A =

P (B) = Probability of choosing box B =

P (C) = Probability of choosing box C =

= The probability of getting 1 blue ball from box A =

= The probability of getting 1 blue ball from box B =

= The probability of getting 1 blue ball from box C =

Applying the Bay's theorem,

Required Probability =
= =
Question 16
1 / -0

In a class of 90 students, 60 play cricket, 20 play hockey and 20 do not play either game. If a student is chosen at random from this class, what is the probability that he plays exactly one game?

A

B

C

D

Solution

Students in class = 90
Student play cricket n(c) = 60
Student play hockey n(h) = 20
Student play none of game = 20
Students in class play games = 90 - 20 = 70
Student plays both games = 60 + 20 - 70 = 10
Student play only cricket = 60 - 10 = 50
Student play only hockey = 20 - 10 = 10
Thus we have select from 50 + 10
So, Required probability =
Question 17
1 / -0

A deck of cards is divided into 2 small decks with unequal number of cards in each deck. The probability of selecting a red ace from the 1^st deck is given to beand that of selecting a black card from the 2^nd deck is . It is also known that the 2^nd deck does not have any red ace.
Find out the total number of cards in the 2^nd deck.

A
27

B
29

C
30

D
32

Solution

Since, there is no red ace in the 2^nd deck, both the red ace will be in the 1^st deck only.
Now, P(getting a red ace from 1^st deck) =
Let the total number of cards in the 1^st deck = x
So, =
⇒x = 22
So, number of cards in the 1^st deck = 22
Number of cards in 2^nd deck = 52 - 22 = 30
Question 18
1 / -0

John, Mathew and Adam are playing with a pair of dice and have to obtain the sum of numbers 5, 6 and 7, respectively. Which of them has the highest possibility of winning?

A
John

B
Mathew

C
Adam

D
All have equal possibility

Solution

Probability of getting 5 is (2, 3), (3, 2), (1, 4) and (4, 1) i.e. 4/36 = 1/9
Probability of getting 6 is (2, 4), (4, 2), (3, 3), (5, 1) and (1, 5) i.e. 5/36
Similarly, probability of getting 7 is 6/36 = 1/6.
Thus, this is the highest amongst all of them.
Hence, Adam has the highest possibility of winning.
Question 19
1 / -0

A, B and C are three athletes participating in an event. Probabilities of A, B and C winning are 3/4, 1/3 and 2/5, respectively. What is the probability that at least one of them wins?

A

B

C

D

Solution

P(A) =
P(B) = ; P(C) =
=
= ;
P(at least one of them wins) = 1 - probability that none of them wins
=
Question 20
1 / -0

In a bolt factory, machines A, B, and C manufacture 25%, 35% and 40% of the total output, respectively. Out of the total outputs of A, B and C, 5%, 4% and 2%, respectively, are defective bolts. A bolt is drawn at random and is found to be defective. What is the probability that it is manufactured by machine B?

A

B

C

D

Solution

Probability of getting a defective bolt, which is manufactured by machine B, P(D_B) =×=
Probability of getting a defective bolt P(D) = × + × + × Required probability = = = =
Question 21
1 / -0

At an annual function of a company, 50 employees reached on time. Each employee was given a chit numbered 1-50 to select one awardee for the punctuality price.

Find the probability that the number on the chit of the winner is
(a) a composite number
(b) an odd multiple of 3
(c) has digit '4' in it
(d) neither a prime number nor an even number

A
a , b , c , d

B
a , b , c , d

C
a , b , c , d

D
a , b , c , d

Solution

Sample space = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50)

(a) Outcomes of a composite number = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50
Total outcomes = 34
Probability of that the number on the chit of the winner is a composite number = =
(b) Outcomes of odd multiples of 3 = 3, 9, 15, 21, 27, 33, 39, 45
Total outcomes = 8
Probability that the number on the chit of the winner is an odd multiple of 3 = =

(c) Outcomes having digit '4' = 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49
Total outcomes = 14
Probability that the number on the chit of the winner has 4 in it = =

(d) Not even numbers = odd numbers
Odd numbers from 1 to 50 = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49
Outcomes of neither a prime nor an even from 1 to 50 = 9, 15, 21, 25, 27, 33, 35, 39, 45, 49
Total outcomes = 10
Probability that the number on the chit of the winner has neither a prime number nor an even number = =
a , b , c , d
Question 22
1 / -0

Rahul is playing a game of darts. What is the probability that his dart will go on the inner circle provided that the area of the entire dart is 616 cm² and the radius of the inner circle is half the radius of the dart board? (Use pi = )

A

B

C

D

Solution

Area of the dart = R²=616 cm²

× R² = 616

R² = 616 ×
R² = 196

R = 14 cm

Now,
Radius of the inner circle, r = 14 ÷ 2 = 7 cm

Area of the inner circle = r²
Area = × 7 × 7 = 154 cm²Probability of hitting the inner circle = =

Question 23

1 / -0

Match the following.

Column I	Column II
1. If two dices are tossed, then the probability that the sum of the numbers on the dice is a multiple of 4 is
2. If three coins are tossed, then the probability of getting tails on the first coin is
3. If a card is drawn from a pack of 52 cards, then the probability of getting a nine of red colour is
4. If a letter is chosen at random from the word BIBLIOGRAPHY, then the probability of getting a consonant is

1 , 2 , 3 , 4

Solution

Probability =

1. Total number of outcomes when two dice are rolled = 36
(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)
Outcomes of getting a sum of multiple of 4 = (1,3), (2,2), (3,1), (2,6), (3,5), (5,3), (4,4), (6,2), (6,6)
Total number of outcomes = 9
Probability = =

2. Total number of outcomes of when 3 coins are tossed = HHH, HHT, HTH, HTT, THH, THT, TTH, TTT = 8
Total number of outcomes of getting tails on the first coin = THH, THT, TTH, TTT = 4

Probability =

3. Total number of outcomes = 52
Probability of getting a nine of red colour =
4. Total number of letters in 'BIBLIOGRAPHY' = 12
Number of consonants = B, B, L, G, R, P, H, Y = 8

Probability = =

Hence, option 2 is correct.

Question 24

1 / -0

Three bags contain 18 balls of red and blue colours. The first bag contains 3 red and 3 blue balls. The second bag contains 4 red and 2 blue balls and the third bag contains 2 red and 4 blue balls. One ball is taken out at random from each bag. The probability of getting 2 red balls and 1 blue ball is

Solution

The possible cases of taking out 2 red and 1 blue balls are:

Bag	Red balls	Blue balls	Total balls
1	3	3	6
2	4	2	6
3	2	4	6

P(Red, red, blue) =
P(Red, blue, red) =
P(Blue, red, red) =
So, required probability =

Question 25
1 / -0

Fill in the blanks.

1. If three coins are tossed together, the probability of getting tails on the third coin is M.
2. The probability of getting a sum of multiple of 5 if two dice are thrown is N.
3. In a stationery shop, there are 3 pastel colours, 2 painting colours and 5 water colours. The probability of not picking a painting colour is O.
4. If two dice are thrown simultaneously, the probability of getting same prime numbers on both the dice is P.

A
M = , N = , O = , P =

B
M = , N = , O = , P =

C
M = , N = , O = , P =

D
M = , N = , O = , P =

Solution

(1) Total number of outcomes when 3 coins are tossed = HHH, HHT, HTH, HTT, THH, THT, TTH, TTT = 8
Number of outcomes of getting tails on the third coin = HHT, HTT, THT, TTT = 4
Probability of getting tails on the third coin = =

(2) Number of outcomes when two dice are rolled =36
(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)
Number of outcomes of getting a sum of multiples of 5 = (1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4) = 7
Probability of getting a sum of multiples of 5 =

(3) Total numbers of colours = 10
Probability of picking a painting colour = =
Probability of not picking a painting colour = 1 - =

(4) Number of outcomes when two dice are rolled =36
(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)

Outcomes of getting same prime numbers on both the dice = (2,2), (3,3), (5,5)
Probability of getting same prime numbers on both the dice = =
M = , N = , O = , P =