Probability Test

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Probability Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

A natural number is selected at random and is divided by 4. What is the probability that the resulting number will be an integer?

A

B

C

D

Solution

For the resulting number to be an integer, it should be a multiple of 4.
Every 4^th number from 1 is divisible by 4.
Required probability =
Question 2
1 / -0

Three dice are rolled together. What is the probability of getting the same number on the three?

A

B

C

D
None of these

Solution

Total number of possible cases = 6³ = 216
Favourable number of cases = 6
Probability =
Question 3
1 / -0

If four coins are tossed, then what are the chances of getting two heads and two tails?

A

B

C

D

Solution

Total number of outcomes = 2⁴ = 16

Number of favourable outcomes = (HHTT), (HTHT), (HTTH), (TTHH), (THTH) and (THHT) = 6

Probability = =
Question 4
1 / -0

One card is drawn from a deck of 52 cards. The probability of drawing a card of club is

A

B

C

D

Solution

Total number of cases = 52
Total number of favourable cases = 13 (As there are total 13 cards of club)
Required probability = =
Question 5
1 / -0

The probability that a card drawn from a pack of 52 cards will be either a diamond or a king is

A

B

C

D

Solution

Total ways = 52
There are 13 diamonds and 4 kings, out of which there is one king in diamonds.
Number of favourable ways = (13 + 4 - 1) = 16
Required probability =
Question 6
1 / -0

A pair of dice is thrown. Find the probability of the numbers appearing whose sum is greater than or equal to 10.

A

B

C

D

Solution

Total number of possible cases = 36

Favourable events = (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6) = 6

Required probability =
Question 7
1 / -0

Two dice are thrown simultaneously. What is the probability of getting two numbers whose sum is odd?

A
1/2

B
17/36

C
1/4

D
19/36

Solution

S = Sample space
n(S) = 36
E = Event of getting two numbers whose sum is odd = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}.
n(E) = 18
P(E) = n(E)/n(S) = 18/36 = 1/2
Question 8
1 / -0

A card is drawn from a well-shuffled pack of 52 cards. The probability of getting a queen of club or king of heart is

A

B

C

D
None of these

Solution

Number of kings of hearts = 1
Number of queen of clubs = 1

Number of favourable events = 2
Total number of outcomes = 52
Required probability =
Question 9
1 / -0

Two dice are rolled one after the other. The probability that the number on the first is smaller than the number on the second is

A

B

C

D

Solution

Total number of cases = 6² = 36

Favourable cases: (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

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

(3, 4), (3, 5), (3, 6)

(4, 5), (4, 6), and (5, 6), i.e. 15.
Required probability = =
Question 10
1 / -0

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

A

B

C

D

Solution

Favourable cases = 3, 6, 9, 12, 15, 18, 7 and 14
∴ Probability =
Question 11
1 / -0

Two dice are rolled and it is found that the sum of the numbers appeared on them is an odd number. What is the probability that both the dice give odd numbers?

A
0

B

C

D

Solution

If the sum is an odd number, both the numbers cannot be odd. (odd + odd = even)
Question 12
1 / -0

A box contains 6 red, 7 green and 5 blue balls. Each ball is of a different size. The probability that the selected ball is the smallest red ball is

A
1/6

B
1/12

C
1/9

D
1/18

Solution

We need to select the smallest red ball, which will be only 1 red ball among the six red balls, as each ball is of a different size.
Now, total balls are 18 in number. So, the probability that the selected ball is the smallest red ball is .
Question 13
1 / -0

If two dice are rolled, find the probability that the sum of the two numbers obtained is a prime number.

A

B

C

D

Solution

When two dice are rolled, there are 36 cases.
Favourable cases in which the sum of numbers appearing is 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)}, i.e. 15 cases
Required probability =
Question 14
1 / -0

A letter is randomly selected from the word "EDUCATION". What is the probability that the letter is a vowel?

A

B

C

D

Solution

In the word "EDUCATION", there are a total 9 letters and 5 vowels (E, U, A, I and O).

Probability of vowel = =
Question 15
1 / -0

A work is given separately to 6 men. Their probabilities of doing the work are , and . Find the probability that the work will be done.

A

B

C

D

Solution

Probability that the work will be done = 1 - Probability that work will not be done

= 1 - {(1 - )(1 - )(1 - )(1 - )(1 - )(1 - )}

= 1 -

= 1 -
=
Question 16
1 / -0

A basket contains seven apples and fourteen oranges. If we select two fruits at random, then what is the probability that both will be different?

A

B

C

D

Solution

Case (1): Event A: First is apple; second is orange. Then, P (A) = (7/21) (14/20) = 7/30.
Case (2): Event B: First is orange; second is apple. Then, P (B) = (14/21) (7/20) = 7/30.
So, the required probability is P (A) + P (B) = 7/15.
Question 17
1 / -0

The probabilities of three doctors, A, B and C, achieving success in an operation are 0.5, 0.2 and 0.3, respectively. Find the probability that the operation is not successful.

A
0.78

B
0.64

C
0.56

D
0.28

Solution

Since A, B or C could perform the operation independently, these are mutually exclusive events.

Therefore, the required probability is (1 - 0.5) (1 - 0.2) (1 - 0.3) = 0.28.
Question 18
1 / -0

While doing net practice, a spinner finds that the probability of his ball hitting the wickets is 1/3. Three balls are bowled simultaneously by him. Find the probability that the wickets will be hit at least once.

A

B

C

D

Solution

Probability of ball not hitting the target =
Probability that none of the 3 balls hit the target =
Probability that at least one of the balls hits the target =
Question 19
1 / -0

The probability that Anu would be alive in 2050 is and that of her husband being alive is . What is the probability that their kids will have at least one of the parents alive in 2050?

A

B

C

D

Solution

P(A) = Probability that Anu is alive = 1/7, P(A') =
P(B) = Probability that her husband is alive = 1/6, P(B') =
Probability that their kids will have at least one of the parents alive = 1 - (Probability of both being not alive)
= 1 - ( x ) =
Question 20
1 / -0

A man and his wife appear in an interview for two vacancies in the same post. The probability of the husband's selection is and that of the wife's selection is . What is the probability that neither of them will be selected?

A

B

C

D

Solution

Required probability =
Question 21
1 / -0

State True (T) and False (F) for the following statements.

(i) The probability of an event can never be greater than 1.
(ii) When two dice are rolled, the probability of getting the sum of 13 of two numbers appeared is .
(iii) In a 12-month period, the probability of a month having 31 days is .
(iv) In the English alphabet, if a letter is randomly chosen, the probability that the letter is a consonant is .

A
(i) - T, (ii) - T, (iii) - F, (iv) - T

B
(i) - T, (ii) - F, (iii) - F, (iv) - T

C
(i) - T, (ii) - T, (iii) - T, (iv) - F

D
(i) - F, (ii) - T, (iii) - F, (iv) - T

Solution

(i) The probability of an event can never be greater than 1. This statement is true as the probability of an event lies between 0 and 1 (both inclusive).
(ii) When two dice are rolled, the probability of getting the sum of 13 of two numbers appeared is . This statement is false because when two dice are rolled, the maximum number as the sum of both the numbers on the dice is 12, i.e. 6 + 6. Therefore, this is an impossible event.
(iii) In a 12-month period, the probability of a month having 31 days is . This statement is also false as there are seven months having 31 days in a year.
So, probability of a month having 31 days =
(iv) In the English alphabet, if a letter is randomly chosen, the probability that the letter is a consonant is . This statement is true as there are 21 consonants and 5 vowels in the English alphabet.
Question 22
1 / -0

A card is drawn from a well shuffled pack of cards. Which of the following statements is/are CORRECT in this regard?

Statement-1: The probability of the selected card being a diamonds or a hearts is .
Statement-2: The probability of the selected card being a red coloured bearing an even number is .

A
Only Statement-1

B
Only Statement-2

C
Both Statement-1 and Statement-2

D
Neither Statement-1 nor Statement-2

Solution

Statement-1 : A well shuffled pack of cards has 52 cards in it, having 13 each of spades, diamonds, hearts and clubs.
Number of card of diamonds or hearts = 26
Probability of the selected card being a diamonds or hearts = = 26/52 = 1/2
Number of red coloured cards = 26

Statement-2 : Number of even numbered red coloured cards = 10 [5 each of hearts and diamonds (2, 4, 6, 8, 10)]

Probability of the selected card being a red coloured bearing an even number = = 10/52 = 5/26

Question 23

1 / -0

Two bats are flipped simultaneously. Find A and B, respectively.

Number of hills up		Required probability
(i)	0	A
(ii)	2	B

1/2, 1/2

1/2, 1/4

1/4, 1/4

1/4, 1/2

Solution

Let hills up be denoted by H and flats up be denoted by F.
Total number of outcomes = {HH, HF, FH, FF} (4 outcomes)
From the given information,
(i) Probability of no hills up = 1/4 (only one possibility out of 4)
(ii) Probability of two hills up = 1/4 (only one possibility out of 4)

Question 24

1 / -0

The following table gives the estimated height of people living in a society.

Number of people	Height (in inches)
13	48-51
9	51-54
23	54-57
10	57-60
5	60-63

Note: 48-51 means (48 ≤ height < 51)

If a person is selected at random from the society, then which of the following statements is INCORRECT?

The probability of the persons having height greater than or equal to 51 inches is 47/60.

The probability of the persons having height less than 60 inches is 5/6.

The probability of the persons having height less than 5 feet 3 inches is 1.

The probability of the persons having height greater than or equal to 4 feet 6 inches is 19/30.

Solution

Number of persons having height less than 60 inches = Number of persons having height between 48 inches and 60 inches = 13 + 9 + 23 + 10 = 55
Probability of the persons having height less than 60 inches = 55/60 = 11/12

Question 25

1 / -0

A survey was conducted on 500 students by a certain group to analyse the sectors which they chose after passing class 12^th, having their stream of education as Arts, Commerce and Science after class 10^th. The data of the survey is given below.

	Arts	Commerce	Science
Bachelor's Degree	50	100	210
Business	2	30	20
Sports	4	6	3
Photography	16	4	1
Others	28	10	16

Find the probability of students:
(a) Having Science or Arts as their stream choosing the Business sector
(b) Having Photography as their sector
(c) Having Commerce as their stream choosing the Sports or 'Others' sector with respect to all the Commerce students.

	(a)	(b)	(c)
(i)	17/350	19/500	8/75
(ii)	11/175	21/500	8/75
(iii)	11/350	21/500	8/75
(iv)	11/175	21/500	11/75

(i)

(ii)

(iii)

(iv)

Solution

Total number of students = 500
(a) From the table,
Number of students choosing the Business sector having Science or Arts as their stream of education = 20 + 2 = 22
Total number of students having Science or Arts as their stream = 50 + 2 + 4 + 16 + 28 + 210 + 20 + 3 + 1 + 16 = 350
So, the probability of students having Science or Arts as their stream choosing the Business sector = 22/350 = 11/175

(b) From the table,
Number of students choosing Photography as their sector = 16 + 4 + 1 = 21
So, the probability of students having Photography as their sector = 21/500

(c) From the table,
Number of Commerce students = 100 + 30 + 6 + 4 + 10 = 150
Number of students having Commerce as their stream choosing the Sports or 'Others' sector = 16
So, the probability of students having Commerce as their stream choosing the Sports or 'Others' sector with respect to all the Commerce students = 16/150 = 8/75

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Probability Test - 7

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Probability Test - 7

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SHARING IS CARING

Question 1

Solution

Question 2

None of these

Solution

Question 3

Solution

Question 4

Solution

Question 5

Solution

Question 6

Solution

Question 7

1/2

17/36

1/4

19/36

Solution

Question 8

None of these

Solution

Question 9

Solution

Question 10

Solution

Question 11

0

Solution

Question 12

1/6

1/12

1/9

1/18

Solution

Question 13

Solution

Question 14

Solution

Question 15

Solution

Question 16

Solution

Question 17

0.78

0.64

0.56

0.28

Solution

Question 18

Solution

Question 19

Solution

Question 20

Solution

Question 21

(i) - T, (ii) - T, (iii) - F, (iv) - T

(i) - T, (ii) - F, (iii) - F, (iv) - T

(i) - T, (ii) - T, (iii) - T, (iv) - F

(i) - F, (ii) - T, (iii) - F, (iv) - T

Solution

Question 22

Only Statement-1

Only Statement-2

Both Statement-1 and Statement-2

Neither Statement-1 nor Statement-2

Solution

Question 23

1/2, 1/2

1/2, 1/4

1/4, 1/4

1/4, 1/2