JEE Mathematics Probability MCQs Set 30

Answer: (b) \( \frac25 \)

Explanation:
1. If the first ball is red (\( \frac4{10} \)), the bag has 6 red and 6 black, so red has probability \( \frac6{12} \).
2. If the first is black (\( \frac6{10} \)), the bag has 4 red and 8 black, so red has probability \( \frac4{12} \).
3. \( P=\frac4{10}\cdot\frac6{12}+\frac6{10}\cdot\frac4{12}=\frac{48}{120}=\frac25 \). Answer: (b)

Teacher's Note:
1. Condition on the first draw.
2. Update the bag contents.

Answer: (c) 2

Explanation:
1. The standard deviation does not change on shifting every observation by 5, so use \( y_i=x_i-5 \).
2. \( \sigma^2=\frac{45}9-\left(\frac99\right)^2=5-1=4 \).
3. \( \sigma=2 \). Answer: (c)

Teacher's Note:
1. Shifting leaves the spread unchanged.
2. Use variance \( =\overline{y^2}-\bar y^2 \).

Answer: (b) \( \frac7{16} \)

Explanation:
1. \( P(\text{exactly one of }A,B)=P(A)+P(B)-2P(A\cap B)=\frac14 \), and similarly for the other two pairs.
2. Adding: \( 2[P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)]=\frac34 \), so the bracket is \( \frac38 \).
3. \( P(A\cup B\cup C)=\frac38+P(A\cap B\cap C)=\frac38+\frac1{16}=\frac7{16} \). Answer: (b)

Teacher's Note:
1. Write each condition in terms of \( P(A),P(A\cap B) \).
2. Add the three equations.

Answer: (a) \( \frac6{55} \)

Explanation:
1. There are \( {}^{11}C_2=55 \) pairs. If \( x+y=4m \) and \( x-y=4n \), then \( x=2(m+n) \) and \( y=2(m-n) \) are both even.
2. Checking even numbers: \( (0,4),(0,8),(2,6),(2,10),(4,8),(6,10) \), which is 6 pairs.
3. Probability \( =\frac6{55} \). Answer: (a)

Teacher's Note:
1. Show both numbers must be even.
2. List pairs of even numbers.

Answer: (b) \( 3a^2-32a+84=0 \)

Explanation:
1. \( (3.5)^2=\frac{4+9+a^2+121}4-\left(\frac{16+a}4\right)^2 \), i.e. \( \frac{49}4=\frac{134+a^2}4-\frac{(16+a)^2}{16} \).
2. Multiply by 16: \( 196=536+4a^2-256-32a-a^2 \), so \( 3a^2-32a+84=0 \). Answer: (b)

Teacher's Note:
1. Write variance as mean of squares minus square of mean.
2. Clear the denominators.

Answer: (d) 14.0

Explanation:
1. The original sum is \( 16\times16=256 \).
2. New sum \( =256-16+(3+4+5)=252 \) with \( 18 \) observations.
3. New mean \( =\frac{252}{18}=14 \). Answer: (d)

Teacher's Note:
1. Find the old sum.
2. Adjust for the deleted and added values.

Answer: (a) independent but not equally likely

Explanation:
1. \( P(A\cup B)=\frac56 \) and \( P(A)=\frac34 \).
2. \( \frac56=\frac34+P(B)-\frac14\Rightarrow P(B)=\frac13 \).
3. \( P(A)P(B)=\frac34\cdot\frac13=\frac14=P(A\cap B) \), so they are independent, and \( P(A)\ne P(B) \). Answer: (a)

Teacher's Note:
1. Find \( P(A\cup B) \) and \( P(A) \).
2. Compare \( P(A\cap B) \) with \( P(A)P(B) \).

Answer: (b) 833

Explanation:
1. The numbers are \( 2,4,\ldots,100 \), with mean \( \frac{50\cdot51}{50}=51 \).
2. \( \overline{x^2}=\frac{4(1^2+\cdots+50^2)}{50}=\frac4{50}\cdot\frac{50\cdot51\cdot101}6=3434 \).
3. Variance \( =3434-51^2=3434-2601=833 \). Answer: (b)

Teacher's Note:
1. Find the mean.
2. Use the sum of squares formula.

Answer: (c) \( \frac{11}{3^5} \)

Explanation:
1. \( p=\frac13 \), \( q=\frac23 \), \( n=5 \).
2. \( P(X\ge4)={}^5C_4\left(\frac13\right)^4\frac23+\left(\frac13\right)^5=\frac{10}{3^5}+\frac1{3^5}=\frac{11}{3^5} \). Answer: (c)

Teacher's Note:
1. Use the binomial distribution.
2. Add \( P(4) \) and \( P(5) \).

Answer: (c) A is true, but R is false.

Explanation:
1. Multiplying each observation by 2 multiplies the variance by 4, so the Assertion is true.
2. The mean of the new data is \( 2\bar x \), not \( \bar x \), so the Reason is false. Answer: (c)

Teacher's Note:
1. Scale the observations by 2.
2. Find the new mean.

Answer: (b) \( \left[0,\frac12\right] \)

Explanation:
1. \( P(\text{at least one failure})=1-p^5\ge\frac{31}{32}\Rightarrow p^5\le\frac1{32} \).
2. So \( p\le\frac12 \), and since \( p\ge0 \), \( p\in\left[0,\frac12\right] \). Answer: (b)

Teacher's Note:
1. Use the complement: no failure means all successes.
2. Take the fifth root.

Answer: (a) \( P(C/D)\ge P(C) \)

Explanation:
1. \( C\cap D=C \), so \( P(C/D)=\frac{P(C)}{P(D)} \).
2. Since \( P(D)\le1 \), \( \frac{P(C)}{P(D)}\ge P(C) \). Answer: (a)

Teacher's Note:
1. Simplify \( C\cap D \).
2. Use \( P(D)\le1 \).

Answer: (b) 4

Explanation:
1. The median of \( a,2a,\ldots,50a \) is \( 25.5a \).
2. Mean deviation \( =\frac{|a|}{50}\cdot2(0.5+1.5+\cdots+24.5)=\frac{|a|}{50}\cdot2\cdot312.5=12.5|a| \).
3. \( 12.5|a|=50\Rightarrow|a|=4 \). Answer: (b)

Teacher's Note:
1. Find the median.
2. Sum the absolute deviations.

Answer: (a) \( P(A^c)-P(B) \)

Explanation:
1. \( P(A^c\cap B^c\cap C)=P(C)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C) \).
2. By pairwise independence this equals \( P(C)[1-P(A)-P(B)] \), so dividing by \( P(C) \) gives \( 1-P(A)-P(B)=P(A^c)-P(B) \). Answer: (a)

Teacher's Note:
1. Expand using inclusion and exclusion.
2. Use \( P(A\cap C)=P(A)P(C) \).

Answer: (b) \( \frac{11}2 \)

Explanation:
1. \( \sum x^2=5(4+4)=40 \) and \( \sum y^2=5(5+16)=105 \). Combined mean \( =3 \).
2. Variance \( =\frac{40+105}{10}-9=\frac{145-90}{10}=\frac{11}2 \). Answer: (b)

Teacher's Note:
1. Find the sums of squares.
2. Use the combined mean.

Answer: (c) A is true, but R is false.

Explanation:
1. Number of APs with common difference \( 1,2,3,4,5,6 \) is \( 17,14,11,8,5,2 \), total \( 57 \), and \( {}^{20}C_4=4845 \).
2. \( \frac{57}{4845}=\frac1{85} \), so the Assertion is true.
3. Common difference 6 is also possible (for example 1, 7, 13, 19), so the Reason is false. Answer: (c)

Teacher's Note:
1. Count APs for each common difference.
2. Check whether \( d=6 \) is possible.

Answer: (c) 10.1

Explanation:
1. The mean is \( 1+50d \). The deviations are \( |kd-50d| \) for \( k=0,\ldots,100 \).
2. Sum \( =2d(1+2+\cdots+50)=2550d \), so mean deviation \( =\frac{2550d}{101}=255 \).
3. \( d=\frac{255\cdot101}{2550}=10.1 \). Answer: (c)

Teacher's Note:
1. Find the mean.
2. Sum the absolute deviations symmetrically.

Answer: (a) \( \dfrac1{\log_{10}4-\log_{10}3} \)

Explanation:
1. \( 1-\left(\frac34\right)^n\ge\frac9{10}\Rightarrow\left(\frac43\right)^n\ge10 \).
2. Taking \( \log_{10} \): \( n(\log_{10}4-\log_{10}3)\ge1 \), so \( n\ge\frac1{\log_{10}4-\log_{10}3} \). Answer: (a)

Teacher's Note:
1. Use the complement.
2. Take logarithms to base 10.

Answer: (a) \( \frac1{14} \)

Explanation:
1. Product zero: tickets 00-09 (10) and 10, 20, 30, 40 (4), so 14 tickets.
2. Among them the digit sum is 8 only for 08, giving 1 ticket.
3. \( P=\frac{1}{14} \). Answer: (a)

Teacher's Note:
1. List tickets with a zero digit.
2. Pick those with digit sum 8.

Answer: (c) \( \frac13 \)

Explanation:
1. \( P(A\cap B)=P(B/A)P(A)=\frac23\cdot\frac14=\frac16 \).
2. \( P(B)=\frac{P(A\cap B)}{P(A/B)}=\frac{1/6}{1/2}=\frac13 \). Answer: (c)

Teacher's Note:
1. Find \( P(A\cap B) \).
2. Divide by \( P(A/B) \).

Answer: (a) \( a=3,b=4 \)

Explanation:
1. Mean: \( a+b+23=30\Rightarrow a+b=7 \).
2. Variance: \( (6-a)^2+(6-b)^2+4+1+16=34 \), so \( (6-a)^2+(6-b)^2=13 \).
3. \( a=3,b=4 \) gives \( 9+4=13 \). Only \( a=3,b=4 \) among the options satisfies both conditions (for \( a=5,b=2 \) the squares sum is 2 and for \( a=1,b=6 \) it is 25). Answer: (a)

Teacher's Note:
1. Use the mean for \( a+b \).
2. Use the variance for the squares of deviations.

Answer: (d) \( 8/243 \)

Explanation:
1. A score of 9 on a single throw has probability \( \frac4{36}=\frac19 \).
2. \( P=3\cdot\left(\frac19\right)^2\cdot\frac89=\frac{24}{729}=\frac8{243} \). Answer: (d)

Teacher's Note:
1. Find the single-throw probability.
2. Use the binomial formula.