Probability MCQ Quiz - Objective Question with Answer for Probability - Download Free PDF

Given:

 A shelf in a library has nine books of three different subjects. Out of these, two are of physics, four are of chemistry and the rest are English

Calculation:

Total outcome = 9!/(2! × 4! × 3!) = 1260

Considering all the English books as one unit, we have 2 physics, 4 chemistry and one unit of English = 7

⇒ Now we can arrange all of these in 7!/(2! × 4!) = 105

Given:

We are selecting a leap year at random.

We need the probability that it has either 53 Tuesdays or 53 Wednesdays.

Calculation:

A leap year has 366 days = 52 weeks + 2 extra days.

The extra 2 days can be: (Sun-Mon), (Mon-Tue), (Tue-Wed), (Wed-Thu), (Thu-Fri), (Fri-Sat), or (Sat-Sun)

Favorable outcomes:

53 Tuesdays occur if extra days are: (Mon-Tue) or (Tue-Wed)

53 Wednesdays occur if extra days are: (Tue-Wed) or (Wed-Thu)

So, favorable pairs = (Mon-Tue), (Tue-Wed), (Wed-Thu)

Total possible 2-day combinations = 7

Favorable cases = 3

∴ Required probability = 3 / 7

Answer: Option (3)

Concept:

Probability = Possible number of outcomes/total number of outcomes

Calculation:

Total number of outcomes = 12

Possible number of outcomes = Number of tickets having a number as a multiple of 2 or 3 = {2, 3, 4, 6, 8, 9, 10, 12} = 8

∴ Probability = 8/12 = 2/3

Hence, the required probability is \(\frac{2}{3}\).

Given :

P(A̅) = \(\frac{2}{9}\) 

Calculation:

We know that,

P(A) + P(A̅) = 1

⇒ P(A) = 1 - P(A̅)

⇒ P(A) = 1 - \(\dfrac{2}{9}\) = \(\dfrac{7}{9}\)

P(A) ∶ P(A̅) = \(\dfrac{7}{9}\) : \(\dfrac{2}{9}\)  = 7 : 2

∴ The answer is 7 : 2.

Given:

Probability of A getting selected = ²/₅

Probability of B getting selected = ⁴/₇

Formula Used:

Probability of both A and B getting selected = Probability of A getting selected × Probability of B getting selected

Calculation:

Probability of both A and B getting selected = ²/₅ × ⁴/₇

⇒ Probability = ^{2 × 4}/_{5 × 7}

⇒ Probability = ⁸/₃₅

The probability that both of them get selected is ⁸/₃₅.

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 = \(\frac 1 2 \times \frac 2 3 \times \frac 3 4\) = 1/4

⇒ P(None) = 1/4

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

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

Given:

Favorable outcomes are a multiple of 2 on one dice and a multiple of 3 on the other dice.

Concept used:

When two dice are thrown total outcome = 6 × 6 = 36

Probability = favourable outcomes/total outcomes

Explanation:

Favourable outcomes = (2,3), (4,3), (6,3), (2,6), (4,6), (6,6),

(3,2), (3,4), (3,6), (6,2), (6,4) = 11

Total outcomes = 6 × 6 = 36

∴ Probability = 11/36

Number of coins tossed = 4

GIVEN:

One dice shows a multiple of 3.

Other dice shows even number.

CONCEPT:

Total number of outcomes in two dice is 36.

FORMULA USED:

P = Favorable outcomes/Total outcomes 

CALCULATION:
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.

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

Number of black balls = 5

Number of white balls = 6

Formula

Probability = Favorable events/Total possible events

Calculation

Favorable event = ⁶C₂

Total possible events = ¹¹C₂

∴ Probability = ⁶C₂/¹¹C₂ = (6 × 5)/(11 × 10) = 3/11

Given:

Three coins are tossed simultaneously.

Formula:

Probability = Number of favorable outcomes/ Total number of outcomes.

Calculation:

When three coins are tossed then the outcome will be any one of these combinations. (TTT, THT, TTH, THH. HTT, HHT, HTH, HHH).

So, the total number of outcomes is 8.

Now, for exactly two heads, the favorable outcome is (THH, HHT, HTH).

We can say that the total number of favorable outcomes is 3.

Again, from the formula

Probability = Number of favorable outcomes/Total number of outcomes

Probability = 3/8

∴ The probability of getting exactly two heads is 3/8.

Calculation

Number of composite number in a dice are  (4 and 6)

⇒ Probabilty of composite number in a dice = 2/6 = 1/3

⇒ Number of prime number in a dice = 2, 3 and 5

⇒ Probability of prime number in a dice = 3/6 = 1/2

∴The probability of getting a composite number on the first roll and a prime number on the second roll = 1/2 × 1/3 = 1/6

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

GIVEN:

Number of unbiased dice = 2

CONCEPT:

Probability (Event) = 1 - (Number of non favorable outcome / Total outcome)

Probability of getting a sum greater than 5 = 1 - (Probability of getting a sum less than or equal to 5)

CALCULATION:

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

Let F = event of getting a sum less than or equal to 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4,1)}

n(F) = 10

⇒ Required probability = 1 - (10/36) = 1 - (5/18) = 13/18

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

Important Points

When we have a large number of cases like 26 in case of Event (E) then we calculate non-favorable outcome(Compliment event i.e. 1 - favourable event)

Mistake Points

In this question, we have to avoid the cases in which sum of digit is equal to five like{(1, 4),(2, 3),(3, 2),(4,1)}

Additional Information

Probabilities for the two dice

Given:

A bag contains 2 red 3 green and 2 blue balls, Two balls are drawn at random. 

Formula Used:

Probability = favourable outcome/total outcome

Calculation:

None of the balls drawn is blue, this can only happen when the two balls drawn at random are either red and green or both.

Total number of balls = 2 + 3 + 2 = 7

⇒ Number of ways of drawing 2 balls out of 7 = ⁷C₂ = (7 × 6) / (2 × 1) = 42/2 = 21

⇒ Number of ways of drawing 2 Blue balls = ²C₂ = 1

So the probability of both balls being blue = 1/21

And the probability of one ball being blue = (²C₁ ×  ⁵C₁)/⁷C₂ = 10/21

Then

⇒ the probability that none of the balls being blue = 1 - 1/21 - 10/21 = 10/21

∴ Required Probability = 10/21