MCQ Test: Probability- 1 - Bank Exams MCQ

Total number of students in class IX = 30

Number of students in club A = 6

Number of students in club B = 7

Number of students in club C = 2

Number of students in club D = 30 - 6 - 7 - 2 = 15

The probability of selecting a student from club A or D is given by:

P(A or D) = P(A) + P(D)

We know that P(A) = Number of students in club A / Total number of students in class IX

So, P(A)= 6/30 = 1/5

Similarly, P(D) = 15/30 = 1/2

Therefore, P(A or D) = P(A) + P(D) = 1/5 + 1/2 = 7/10

Hence, the probability that the selected student is from club A or D is 7/10.

GIVEN:

Number of unbiased dice = 2

CONCEPT:

Probability (Event) = Number of favorable outcome / Total outcome

CALCULATION:

No. of ways of rolling a pair of dice = 6 × 6 = 36

Let E = event of getting a sum greater than 5 = {(1, 6), (1, 5), (2, 6),(2, 5), (2, 4), (3, 6), (3, 5), (3, 4), (3, 3), (4, 6), (4, 5), (4, 4), (4, 3), (4, 2), (5, 6), (5, 5),(5, 4), (5, 3), (5, 2), (5, 1), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6,1)}

n(E) = 26

⇒ Required probability = 26/36 = 13/18

⇒ The probability of getting sum greater than 5 = 13/18

Probability of 3 students,

P(A) = 1/2, P(A̅) = 1/2

P(B) = 1/3, P(B̅) = 2/3

P(C) = 1/4, P(C̅) = 3/4

So, Probability of no one solve the question is

⇒ P(None) = 1/4

Then, The probability to solve the question is = 1 - 1/4 = 3/4

Hence, the correct answer is "3/4".

The total number of students = 18

When 1 name was selected from 18 names, the probability that he was of section B

= 6/18 = 1/3

But from the question, there are 6 students from the section B and the age of all 6 are different therefore, the probability of selecting one i.e. youngest student from 6 students will be 1/6

Hence, option C is correct.

The probability of an even number appearing on the first draw is 1/2( since there are 25 even numbers in counting of 1 to 50).

The probability of an odd number appearing on the second draw is 1/2( since there are 25 odd numbers in counting of 1 to 50).

The probability of a number divisible by 3 appearing on the third draw is 16/50 ( since there are 16 numbers that are divisible by 3 while counting from 1 to 50.)

Since all these events have no relation with each other and no dependence either, and the slips are replaced, we can directly multiply the individual probabilities to get the resultant probability.

So, the probability of all the events taking place is

Hence, option B is correct.

There are 12 face cards in a deck of 52 playing cards, which includes 3 face cards (jack, queen, and king) each for the 4 suits (hearts, diamonds, clubs, and spades). There are 6 red-face cards and 6 black-face cards.

After removing 6 red face cards, the remaining cards are 46.

Remaining face cards 6

The probability of drawing a face card now is 6/46 = 3/23

Therefore, the probability of drawing a face card from the remaining deck is 3/23.

Number of possible combination of 3 persons in which 2 have to be women = (2 Women out of 5 x 1 Man out of 6) or (3 Women out of 5)

= (⁵C₂ x ⁶C₁ + ⁵C₃)

Total possible outcomes = ¹¹C₃

Hence, option B is correct.

Let E₁, E₂ be the event of picking a green bulb and white bulb respectively.

Total no. of bulbs in a bag = 3 + 4 + 5 = 12

Hence, option B is correct.

We fix the red face and to its left pink face and bottom face as blue 

The number of ways to arrange the other three colors = 3!

Total ways of painting the six colors

First we fix any one color on any one face, let’s say red color.

The number of ways five color can be painted = 5!

Eliminating the repeated possibilities

We divide by four to eliminate the repeated possibilities as shown in the figure below. These possibilities are counted as different but don’t give us a different arrangement. The arrangement in all four is same.

Probability (Red, pink and blue share a common corner)

Hence, option D is correct.

Number of ways in which the person can pick three balls out of 18 balls = 18C3 = 816

Number of ways of picking 3 balls of same colour = ⁶C₃+ ⁴C + ⁸C₃ = (20 + 4 + 56) = 80

Probability of picking three balls of same color

=80/816 = 5/51

Required probability = 1 – probability of picking three balls of same colour

Hence, option D is correct.

There are 20 prime numbers between 10 and 90:11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89.

Out of these, the two-digit primes are:  23, 37, 53, 73.

Therefore, there are 4 two-digit primes among the discs numbered from 10 to 90.

The total number of discs in the box is 81 (90 - 10 + 1).

Hence, the probability of drawing a disc with a two-digit prime number is: 4/81.

There are 10 even numbers in the group 1-21.

∴ The probability that the first ball is even numbered = 10/21

Since the ball is not replaced there are now 20 balls left, of which 9 are even numbered.

∴ The probability that the second ball is even numbered = 9/20

Hence, option C is correct.

Number of coins tossed = 4

∴ Sample space of four coins tossed = 2⁴ = 16

Total boxes = 16

Total pieces = 16

Similar pieces = 8 pawns, 2 bishops, 2 rooks, 2 knights

Total ways of arranging these 16 pieces in 16 boxes

Ways of correct arrangement = 1

Hence, option B is correct..

There are 31 days in July, out of which generally at least 4 days are Mondays.

But as months starts with Monday, it must have 5 Mondays.

Therefore, there are 26 days that are not Mondays. The probability of selecting a day that is not a Monday is 26/31.

Hence, the probability that a day selected at random from July is not a Monday is 26/31.

Let, the number of pink balls be p

Probability of choosing a pink ball = P/35

So, remaining number of balls = (35 – 15) = 20

Hence, option A is correct.

The perfect squares among the numbers from 1 to 6 are 1 and 4.

The total number of outcomes when the dice is rolled once is 6.

The probability of getting a 1 or 4 is the sum of the frequencies of these outcomes divided by the total number of outcomes, i.e.,

P(getting a perfect square) = (160 + 50)/600

⇒ P = 210/600

⇒ P = 21/60

⇒ P = 7/20

Therefore, the probability of getting a number which is a perfect square is 7/20.

When 4 fair coins are tossed simultaneously, the total number of outcomes is 2⁴ = 16

At least 3 heads implies that one can get either 3 heads or 4 heads.

One can get 3 heads in ⁴C₃ = 4 ways and can get 4 heads in ⁴C₄ = 1 ways.

∴ Total number of favorable outcomes = 4 + 1 = 5

∴ The reqd. probability =  1/4

Hence, option C is correct.

Given

Probability of passing the test by Sita = 60% = 60/100

Probability of passing the test by Gita = 40% = 40/100

Probability of passing the test by Mita = 20% = 20/100

Formula

Probability of not happening even A = 1 - Probability of  happening even A

Probability of happening A and B = Probability of happening A × Probability of happening B

Calculation

Probability of not passing the test by Mita = 1 - Probability of passing the test by Mita

= 1 - (20/100)

= 80/100

Now,

Probability that at Sita and Gita will pass the test and Mita will not = Probability of passing the test by Sita × Probability of passing the test by Gita × Probability of not passing the test by Mita

= (60/100) × (40/100) × (80/100)

= 192/1000

= (192/10)%

= 19.2%

GIVEN:

One dice shows a multiple of 3.

Other dice shows even number.

Total number of outcomes in two dice is 36.

FORMULA USED:

P = Favorable outcomes/Total outcomes 

There are only 6 such cases as required,

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

∴ Required probability = 6/36 = 1/6

∴ The probability is 1/6.