Problems on bags and balls/similar objects MCQ Quiz in मल्याळम - Objective Question with Answer for Problems on bags and balls/similar objects - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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.

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

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

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.

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.