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

The probability, that the first draw gives all white balls and the second draw gives all black balls, is

\(\frac{{ }^6 \mathrm{C}_4}{{ }^{15} \mathrm{C}_4} \times \frac{{ }^9 \mathrm{C}_4}{{ }^{11} \mathrm{C}_4}=\frac{3}{715}\)

Hence option (3) is correct. 

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.

Explanation:

Here, P(A) = 5/6 , P(B) = 4/5 , P(C) = 3/4

Also

P(B) = \(1 - \frac{4}{5} = \frac{1}{5}\)

Now

⇒ \(P ( A\cap \overline{B} \cap C) = P(A).P(B).P(C)\)

= \(\frac{5}{6}\times \frac{1}{5}\times \frac{3}{4} = \frac{1}{8}\)

∴ Option (a) is correct.

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

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

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