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

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 $={\displaystyle \frac{6}{55}}$

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:

${}^n{C}_r=\frac{n!}{r!(n-r)!}$

0! = 1

Calculation:

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

⇒ $\frac{5\times 4\times 3!}{2!\times 3!}$

⇒ $\frac{5\times 4}{2\times 1}$

⇒ 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 = $\frac{{}^3{C}_2+{}^2{C}_2}{{}^5{C}_2}$

⇒ $\frac{3+1}{10}$

⇒ $\frac{4}{10}$

⇒ $\frac{2}{5}$

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

Calculation

$A\to$ Maximum is 6

$B\to$ Minimum is 3

${\displaystyle P(A)=\frac{{}^5{C}_2}{{}^8{C}_3}}$

${\displaystyle P(B)=\frac{{}^5{C}_2}{{}^8{C}_3}}$

${\displaystyle P(A\cap B)=\frac{{}^2{C}_1}{{}^8{C}_3}}$

${\displaystyle P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}=\frac{{}^2{C}_1}{{}^5{C}_2}=\frac{2}{10}=\frac{1}{5}}$

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 $={\displaystyle \frac{2^9\times 220\times 3}{3^{12}}}$

Probability $=55\times {\displaystyle \frac{2^{11}}{3^{11}}}$

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.

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 ${\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}$

Probability of choosing a white ball then a red ball from bag $A$ is given by ${\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}$

So, the probability of getting a white ball and then a red ball from bag $B$ is given by = ${\displaystyle \frac{{\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}}{{\displaystyle \frac{{}^4{C}_1{\times }^2{C}_1}{{}^9{C}_2}}+{\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}}}$

$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{2}{9}}}{{\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{9}}}}={\displaystyle \frac{2\times 7}{18+14}}={\displaystyle \frac{7}{16}}$

Hence option 1 is correct

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 ${\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}$

Probability of choosing a white ball then a red ball from bag $A$ is given by ${\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}$

So, the probability of getting a white ball and then a red ball from bag $B$ is given by = ${\displaystyle \frac{{\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}}{{\displaystyle \frac{{}^4{C}_1{\times }^2{C}_1}{{}^9{C}_2}}+{\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}}}$

$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{2}{9}}}{{\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{9}}}}={\displaystyle \frac{2\times 7}{18+14}}={\displaystyle \frac{7}{16}}$

Hence option 1 is correct

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.

Explanation:

Given:

Urn A contains 2 white and 2 black balls while urn B contains 3 white and 2 black balls. 

⇒ E1 = One white ball transferred from A to B; then urn A contains white = 1 and b = 2 and urn B contains white = 4 and black = 2

⇒ E2 = One black ball transferred from A to B. Then urn A contains white = 2 and black = 1 and urn B contains white = 3 and black = 3

⇒ F = One white ball is drawn from urn B

⇒ $P(F)=P(E_1).P(\frac{F}{E_1})+P(E_2).P(\frac{F}{E_2})$

= $\frac{2}{4}\times \frac{4}{6}+\frac{2}{4}\times \frac{3}{6}=\frac{7}{12}$

∴ Option (b) is correct.