Probability - 1 - Free MCQ Practice Test with solutions, Class 10 The Complete

Given:

Black balls + Yellow balls = 9

Probability of selecting two red balls = 1/7 = Probability of selecting two black balls 

Concept used:

A combination is the choice of r things from a set of n things without replacement and where order does not matter and is given by 

⇒  ⁿC_r = n!/{r! × (n - r)!}

Probability of an event = number of favourable outcomes/total possible outcomes

Calculation:

Let the number of red and black balls be x and y respectively

Then, probability of selecting two red balls = (^xC₂)/(^{(9 + x)}C₂) = 1/7

⇒ x × (x - 1)/(9 + x)(8 + x) = 1/7

⇒ 7x² - 7x = x² + 17x + 72

⇒ 6x² - 24x - 72 = 0 

⇒ x = 6 

∴ Number of red balls = 6

Total number of balls  = 9 + 6 = 15

Now, the probability of selecting two black balls from the bag = (^yC₂)/(¹⁵C₂) = (1/7) × 100/250

⇒ y(y - 1)/(15 × 14) = (1/7) × (2/5)

⇒ y = 4

∴ The number of black ball = 4

∴ The number of yellow coloured ball = 15 - 6 - 4 = 5 balls.  

Formula used:

Probability of occurrence of the event:

P(E) =  n(E)/n(S)

Where,

n(E) = Number of favorable outcome

n(S) = Number of possible outcome

Calculation:

Total no. of outcomes = 1, 2, 3, 4, 5 & 6

Number of possible outcomes n(S) = 6

Number of favorable outcomes n(E) = 1

Using the above formula 

Probability of getting more than 5 = 1/6

∴ The correct answer is 0.16

Bag B1 
Number of blue pen = 3x
Number of ref pen = 2x
Bag B2
Number of Blue pen = 7
Number of Black pen = 8
Number of Green pen = X
Two pens are drawn from B2 randomly and kept in bag B1.

Total Number of pen in Bag B1 = 3x + 2x + 2 = 5x+2
Total Number of pen in Bag B2 = 13 + x
Blue pen is drawn from bag B1 such that the probability of drawing a blue pen is 5/9
Case 1:
If pen kept in B1 from B2 are both red pen.
probability = (number of blue pen)/(total number of ball in B1)
3x/(5x + 2) = 5/9
x = 5
Case 2:
If pen kept in B1 from B2 are both blue pen.
probability = (number of blue pen)/(total number of ball in B1)
(3x + 2)/(5x + 2) = 5/9
x = -4
Case 3:
if pen kept in B1 from B2 one pen is blue and one pen is red.
probability = (number of blue pen)/(total number of ball in B1)
(3x + 1)/(5x + 2) = 5/9
x = 0.5

As number of pen cannot be in decimal and negative Case 1 is valid. 

Number of pen in bag B1 = 5x = 5 × 5 = 25

Number of coins tossed = 4

∴ Sample space of four coins tossed = 2⁴ = 16

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
Alternate Method:

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:

Probability of red ball = 1/3

Probability of white ball = 1/2

Formula Used:

Probability = (Number of successful outcomes/Total number of outcomes)

P(E) = (nE)/(nS), where nE = Number of events and nS = Number of sample space 

Calculation:

Probability of getting green ball = 1 - (1/3 + 1/2)

⇒ 1 - 5/6 = 1/6

According to the question:

If one unit corresponds to 9 green balls then,

6 unit = 6 × 9 = 54

Total no. Of balls = 54

∴ The total number of balls in the bag is 54.

Formula used:

n(A U B) = n(A) + n(B) - n(A ∩ B)

Calculation:

n(A) = 28

n(B) = 32

n(A U B) = 40

n(A ∩ B) = 32 + 28 - 40 = 20

Concept: 

When a dice is thrown once. The total number of outcomes is 6 (1, 2, 3, 4, 5, and 6)

Numbers not divisible by 2 = 3 (1, 3, 5)

Formula used: 

Probability = No of Favourable Outcomes/ Total no of Outcomes

P( no not divisible by 2) = 3/6 

⇒ 1/2 

∴ The required probability is 1/2.

Formula used:

Probability of an event = Number of favorable outcomes/Total outcomes

Calculation:

Numbers on dice not divisible by 3 are {1, 2, 4, 5}

⇒ Number of favorable outcomes = 4

Total possible outcomes = 6

∴ The probability that the number shown on the dice is not divisible by 3 is 4/6 = 2/3 

Formula used: 

Probability (E) = (Number of favorable outcomes)/(Total no. of possible outcomes)

Calculation:

Total number of cards = 52

After removing one black ace, one black queen and all the four jacks,

Number of remaining card = 52 - 7 = 45

Number of queen remaining = 3

Therefore, ³C₁(Selecting one out of remaining 3)times out of  ⁴⁵C₁( Selecting 1 out of 45 items) a queen is picked.

Now,

Probability (E) = ³C₁/⁴⁵C₁ = ​​3/45 = 1/15

∴ The required probability is 1/15.