Problems on bags and balls/similar objects MCQ Quiz in தமிழ் - Objective Question with Answer for Problems on bags and balls/similar objects - இலவச PDF ஐப் பதிவிறக்கவும்

Given:

• Total red balls = 8

• Total white balls = 5

• Total balls = 8 + 5 = 13

• Number of balls drawn = 3

• Condition: 1 red ball and 2 white balls

Concept Used:

To calculate the probability of an event, we use the formula:

Probability = (Number of favorable outcomes) / (Total possible outcomes)

The total number of ways to choose 3 balls from 13 is given by:

Total combinations: C(13, 3) = 13! / (3! × (13 - 3)!)

For favorable outcomes:

Choose 1 red ball out of 8: C(8, 1) = 8

Choose 2 white balls out of 5: C(5, 2) = 5! / (2! × (5 - 2)!)

Calculation:

Total possible outcomes:

C(13, 3) = 13! / (3! × 10!) = (13 × 12 × 11) / (3 × 2 × 1)

⇒ C(13, 3) = 286

Favorable outcomes:

C(8, 1) = 8

C(5, 2) = 5! / (2! × 3!) = (5 × 4) / (2 × 1)

⇒ C(5, 2) = 10

Total favorable outcomes = C(8, 1) × C(5, 2) = 8 × 10 = 80

Probability:

P = Favorable outcomes / Total outcomes

⇒ P = 80 / 286

Simplify:

⇒ P = 40 / 143

Conclusion:

The probability of drawing 1 red ball and 2 white balls is: 40 / 143

∴ The correct answer is Option 2.

Calculation

Let us take this case by case. First, take number \(0\). The second number \(p\) has to be chosen such that \(p-0\) and \(p+0\) are divisible by \(4\). Hence, \(p\) can be either \(4\) or \(8\).

Now, take number \(1\). We find that \(3+1\) is divisible by \(4\) but \(3-1\) is not. Similarly, \(5-1\) is divisible by \(4\) but \(5+1\) is not. You can see that for all odd numbers, no such \(p\) exists, so only even numbers should be checked.

Hence, total number of cases possible are \(2+2+1+1=6\)

Total ways to choose \(2\) numbers from \(\{0,1,2,...,10\}\) are \({ }^{11}C_2 = 55\)

Hence, probability \(=\dfrac{6}{55}\)

Hence option 1 is correct

Given:

A bag containing 3 red and 2 black balls, two balls are drawn at random.

Concept:

Probability: Probability is defined as the possibility of an event happening which is equal to the ratio of the number of favorable outcomes and the total number of outcomes.

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

Formula Used:

\(^nC_r = \frac{n!}{r!(n \ - \ r)!}\)

0! = 1

Calculation:

Number of ways of drawing 2 balls out of (3 + 2) balls = ⁵C₂

⇒ \(\frac{5 × 4 × 3!}{2! × 3!}\)

⇒ \(\frac{5 × 4}{2 × 1}\)

⇒ 5 × 2

⇒ 10

Number of ways of drawing 2 red balls = ³C₂

⇒ \(\frac{3 \times 2!}{2! \times 1!}\)

⇒ 3

Number of ways of drawing 2 black balls = ²C₂

⇒ \(\frac{2!}{2! \times 0!}\)

⇒ 1   

Probability of getting balls of same color. 

Probability = \(\frac{^3C_2 \ + \ ^2C_2}{^5C_2}\)

⇒ \(\frac{3 \ + \ 1}{10}\)

⇒ \(\frac{4}{10}\)

⇒ \(\frac{2}{5}\)

∴ The probability that they are of the same color is 2/5.

Calculation

\(A\rightarrow\) Maximum is 6

\(B\rightarrow\) Minimum is 3

\(\displaystyle P(A)=\cfrac {^5C_2}{^8C_3}\)

\(\displaystyle P(B)=\cfrac {^5C_2}{^8C_3}\)

\(\displaystyle P(A\cap B)=\cfrac {^2C_1}{^8C_3}\)

\(\displaystyle P(\frac{B}{A})=\cfrac{P(A\cap B)}{P(A)}=\cfrac{^2C_1}{^5C_2}=\cfrac{2}{10}=\cfrac{1}{5}\)

Hence option 2 is correct

Calculation

We are assuming that the given balls are different.

\(n(S) = 3^{12}\) since each ball will go in three ways.

We will assume that at least one of the boxes is having exactly 3 balls.

We can select \(3\) balls from \(12\) in \(^{12}C_3 = 220\) ways.

We can select \(1\) bag from \(3\) in \(^{3}C_1 = 3\) ways.

The rest \(9\) balls can go in \(2^9\) ways.

Hence, Probability \(= \dfrac{2^9 \times 220 \times 3}{3^{12}} \)

Probability \(= 55 \times \dfrac{2^{11}}{3^{11}}\)

Hence option 2 is correct

First, let's find out the total number of balls.

Total balls = Red balls + White balls + Yellow balls

Total balls = 5 (red) + 3 (white) + 2 (yellow)

Total balls = 10

Now, let's calculate the percentage of yellow balls:

(Number of yellow balls / Total balls) * 100%

= (2 / 10) * 100%

= 20%

So, 20% of the balls are yellow.

CONCEPT:

Let S be a sample space and E be an event such that n(S) = n, n(E) = m and each outcome is equally likely. Then \(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{m}{n}\)

CALCULATION:

Given: A bag contains 9 white balls and 12 red balls. 

No. ways to draw a white ball from the bag = C (9, 1) = 9

No. of ways to draw a ball from the bag = C (21, 1) = 21

So, probability of the ball drawn from the bag is white in colour = 9/21 = 3/7

Hence, option D is the correct answer.

Answer (1)

Sol.

p(B₁ / R) = \(\frac{p\left(B_{1}\right) \cdot p\left(R / B_{1}\right)}{p(R)}\)

= \(\frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}}=\frac{1}{4}=p\)

p(B3 / G) = \(\frac{p\left(B_{3}\right) \cdot p\left(G / B_{3}\right)}{p(G)}\)

= \(\frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{4}{10}}=\frac{3}{10}=q\)

\(\frac{1}{p}+\frac{1}{q}=4+\frac{10}{3}=\frac{22}{3}\)

Calculation

For the bag \(A\) we can see that there are \(2\) white, \(3\) red and \(2\) black balls. Similarly, from bag \(B\) we have \(4\) white, \(2\) red and \(3\) black balls.

Probability of choosing a white and then a red ball from bag \(B\) is given by \(\dfrac{^4C_1 \times \ ^2C_1}{^9C_2}\)

Probability of choosing a white ball then a red ball from bag \(A\) is given by \(\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}\)

So, the probability of getting a white ball and then a red ball from bag \(B\) is given by = \(\dfrac{\dfrac{^4C_1 \times \ ^2C_1 }{^9C_2}}{\dfrac{^4C_1 \times ^2C_1}{^9C_2}+\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}}\)

\(\Rightarrow\dfrac{\dfrac{2}{9}}{\dfrac{2}{7}+\dfrac{2}{9}} = \dfrac{2 \times 7}{18+14}=\dfrac{7}{16}\)

Hence option 1 is correct