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
Last updated on Mar 11, 2025
Latest Problems on bags and balls/similar objects MCQ Objective Questions
Top Problems on bags and balls/similar objects MCQ Objective Questions
Problems on bags and balls/similar objects Question 1:
If two different numbers are taken from the set \(\{0,1, 2, 3, ......, 10\}\); then the probability that their sum as well as absolute difference are both multiple of \(4\), is.
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 1 Detailed Solution
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.
First Number | Second Number | No. of cases |
0 | 4,8 | 2 |
2 | 6,10 | 2 |
4 | 0,8(However, case of \(0,4\) is already taken) | 1 |
6 | 2,10 (However, case of \(2,6\) is already taken) | 1 |
8 | 0,4(However, both cases have been taken) | 0 |
10 | 2,6(However, both cases have been taken) | 0 |
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
Problems on bags and balls/similar objects Question 2:
From a bag containing 3 red and 2 black balls, two balls are drawn at random. Find the probability that they are of the same color.
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 2 Detailed Solution
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.
Problems on bags and balls/similar objects Question 3:
Three numbers are chosen at random without replacement from \(\{1, 2, 3, \ldots,8\}\). The probability that their minimum is \(3\), given that their maximum is \(6\), is:
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 3 Detailed Solution
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
Problems on bags and balls/similar objects Question 4:
If \(12\) distinct balls are to be placed in \(3\) identical boxes, then the probability that one of the boxes contains exactly \(3\) balls is
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 4 Detailed Solution
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
Problems on bags and balls/similar objects Question 5:
A boy has 5 red balls, 3 white balls, and 2 yellow balls. What percent of balls are yellow?
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 5 Detailed Solution
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.
Problems on bags and balls/similar objects Question 6:
A bag contains 9 white balls and 12 red balls. If one ball is drawn at random from the bag what is the probability the ball drawn is white in colour ?
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 6 Detailed Solution
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.
Problems on bags and balls/similar objects Question 7:
A box \(A\) contains \(2\) white, \(3\) red and \(2\) black balls. Another box \(B\) contains \(4\) white, \(2\) red and \(3\) black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box \(B\) is
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 7 Detailed Solution
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
Problems on bags and balls/similar objects Question 8:
A box \(A\) contains \(2\) white, \(3\) red and \(2\) black balls. Another box \(B\) contains \(4\) white, \(2\) red and \(3\) black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box \(B\) is
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 8 Detailed Solution
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
Problems on bags and balls/similar objects Question 9:
Urn A contains 2 white and 2 black balls while urn B contains 3 white and 2 black balls. One ball is transferred from urn A to urn B and then a ball is drawn out of urn B. What is the probability that the ball is white?
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 9 Detailed Solution
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.
Problems on bags and balls/similar objects Question 10:
A can hit a target 5 times in 6 shots, B can hit 4 times in 5 shots and C can hit 3 times in 4 shots. What is the probability that A and C may hit but B may lose?
Answer (Detailed Solution Below)
Problems on bags and balls/similar objects Question 10 Detailed Solution
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.