Test: Probability - 1 - Grade 10 MCQ

# Test: Probability - 1 - Grade 10 MCQ

Test Description

## 10 Questions MCQ Test - Test: Probability - 1

Test: Probability - 1 for Grade 10 2024 is part of Grade 10 preparation. The Test: Probability - 1 questions and answers have been prepared according to the Grade 10 exam syllabus.The Test: Probability - 1 MCQs are made for Grade 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Probability - 1 below.
Test: Probability - 1 - Question 1

### In a bag, there are some red, black and yellow balls. Sum of black and yellow balls is 9. Probability of selecting two red balls from that bag is 1/7 which is 250% of the probability of selecting two black balls. Find number of yellow balls in that bag if the number of black balls is even.

Detailed Solution for Test: Probability - 1 - Question 1

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

⇒  nCr = 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.

Test: Probability - 1 - Question 2

### What is the probability of getting more than 5 when dice is trown

Detailed Solution for Test: Probability - 1 - Question 2

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

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Test: Probability - 1 - Question 3

### A cotton bag B1 has blue pens and red pens in a ratio 3 : 2. Another cotton bag B2 has 7 blue pens, 8 black pens and X green pens. Two pens are drawn from B2 randomly and kept in bag B1. Now a blue pen is drawn from bag B1 such that the probability of drawing a blue pen is 5/9. If (X = Number of blue pens in Bag B1 initially), then what is the total number of pen in bag B1 initially?

Detailed Solution for Test: Probability - 1 - Question 3

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

Test: Probability - 1 - Question 4

The sample space of four coins tossed together is:

Detailed Solution for Test: Probability - 1 - Question 4

Number of coins tossed = 4

∴ Sample space of four coins tossed = 24 = 16

Test: Probability - 1 - Question 5

Two unbiased dice are rolled simultaneously. Find the probability of getting sum greater than 5.

Detailed Solution for Test: Probability - 1 - Question 5

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)}

Probabilities for the two dice

Test: Probability - 1 - Question 6

A bag contains only red, green and white balls. The probability of selecting a red ball from the bag at random is 1/3 and that of selecting a white ball at random is 1/2. If the bag contains 9 green balls, the total number of balls in the bag is:

Detailed Solution for Test: Probability - 1 - Question 6

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.

Test: Probability - 1 - Question 7

Comprehension:

A has 28 elements, B has 32 elements and (A U B) has 40 elements.

How many elements does A ∩ B have?

Detailed Solution for Test: Probability - 1 - Question 7

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

Test: Probability - 1 - Question 8

A dice is thrown. What is the probability that the number shown on the dice is not divisible by 2?

Detailed Solution for Test: Probability - 1 - Question 8

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.

Test: Probability - 1 - Question 9

A dice is thrown randomly. What is the probability that the number shown on the dice is not divisible by 3?

Detailed Solution for Test: Probability - 1 - Question 9

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

Test: Probability - 1 - Question 10

From a pack of 52 cards, one black ace, one red king, one black queen and all the four jacks were lost. Find the probability that one card picked out of the remaining cards would be a queen.

Detailed Solution for Test: Probability - 1 - Question 10

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.

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