JEE Mathematics Probability MCQs Set 31

Answer: (c) 0.32

Explanation:
1. The first plane misses with probability 0.7 and the second misses with probability 0.8. Planes alternate until a hit.
2. \( P=(0.7)(0.2)+(0.7)(0.8)(0.7)(0.2)+\cdots=0.14\left[1+0.56+0.56^2+\cdots\right] \).
3. \( =\frac{0.14}{1-0.56}=\frac{0.14}{0.44}=\frac7{22}\approx0.32 \). Answer: (c)

Teacher's Note:
1. Write the probability of each round of misses.
2. Sum the geometric series.

Answer: (c) 80%

Explanation:
1. Let there be \( x \) boys and \( y \) girls: \( 52x+42y=50(x+y)\Rightarrow2x=8y\Rightarrow x=4y \).
2. Boys form \( \frac{4y}{5y}\times100=80\% \). Answer: (c)

Teacher's Note:
1. Use the weighted mean.
2. Find the ratio of boys to girls.

Answer: (c) \( \frac6{e^5} \)

Explanation:
1. \( P(X\le1)=P(0)+P(1)=e^{-5}+5e^{-5} \).
2. \( =\frac6{e^5} \). Answer: (c)

Teacher's Note:
1. Use the Poisson formula with mean 5.

Answer: (a) 24.0

Explanation:
1. Using the empirical relation, mode \( =3\,\text{median}-2\,\text{mean} \).
2. \( =66-42=24 \). Answer: (a)

Teacher's Note:
1. Use the empirical relation.

Answer: (c) \( \frac19 \)

Explanation:
1. Each person has 3 choices, so there are \( 3^3=27 \) cases.
2. Favourable cases: all choose house 1, 2 or 3, so 3 cases, and \( P=\frac3{27}=\frac19 \). Answer: (c)

Teacher's Note:
1. Count total and favourable cases.

Answer: (b) \( 1-\frac3{e^2} \)

Explanation:
1. \( P(X\gt1.5)=P(X\ge2)=1-[P(0)+P(1)] \).
2. \( =1-\left(e^{-2}+2e^{-2}\right)=1-\frac3{e^2} \). Answer: (b)

Teacher's Note:
1. X takes integer values.
2. Use the complement.

Answer: (c) I and II

Explanation:
1. Statement I is true: mode can be read from a histogram. Statement II is true: the median changes when every value is scaled.
2. Statement III is false: variance is independent of origin but not of scale. So I and II are correct. Answer: (c)

Teacher's Note:
1. Test each statement separately.

Answer: (c) 2

Explanation:
1. The mean of all \( 2n \) observations is 0.
2. Variance \( =\frac{\sum x^2}{2n}=\frac{2na^2}{2n}=a^2=4 \), so \( |a|=2 \). Answer: (c)

Teacher's Note:
1. Find the mean first.
2. Compute the variance.

Answer: (c) \( \frac7{20} \)

Explanation:
1. They contradict if exactly one speaks the truth.
2. \( P=\frac45\cdot\frac14+\frac15\cdot\frac34=\frac4{20}+\frac3{20}=\frac7{20} \). Answer: (c)

Teacher's Note:
1. Count the two ways of contradicting.

Answer: (b) 0.77

Explanation:
1. \( E=\{2,3,5,7\} \) so \( P(E)=0.23+0.12+0.20+0.07=0.62 \). \( F=\{1,2,3\} \) so \( P(F)=0.15+0.23+0.12=0.50 \).
2. \( E\cap F=\{2,3\} \) with probability \( 0.35 \).
3. \( P(E\cup F)=0.62+0.50-0.35=0.77 \). Answer: (b)

Teacher's Note:
1. List the outcomes in \( E \) and \( F \).
2. Use the addition theorem.

Answer: (d) \( \frac{28}{256} \)

Explanation:
1. \( np=4 \) and \( npq=2 \) give \( q=\frac12 \), \( p=\frac12 \) and \( n=8 \).
2. \( P(X=2)={}^8C_2\left(\frac12\right)^8=\frac{28}{256} \). Answer: (d)

Teacher's Note:
1. Find \( n \) and \( p \).

Answer: (d) remains the same as that of the original set

Explanation:
1. The median of 9 observations is the 5th value in order.
2. Increasing only the largest 4 observations does not change the order or the 5th value, so the median stays 20.5. Answer: (d)

Teacher's Note:
1. Identify the median position.

Answer: (a) 78.00

Explanation:
1. Corrected \( \sum x^2=2830-400+900=3330 \) and \( \sum x=170-20+30=180 \).
2. Variance \( =\frac{3330}{15}-\left(\frac{180}{15}\right)^2=222-144=78 \). Answer: (a)

Teacher's Note:
1. Correct both sums.
2. Apply the variance formula.

Answer: (d) \( \frac25 \)

Explanation:
1. The probability that both selected horses lose is \( \frac45\cdot\frac34=\frac35 \).
2. So the probability that the winner is selected is \( 1-\frac35=\frac25 \). Answer: (d)

Teacher's Note:
1. Use the complement.

Answer: (a) \( \left[\frac13,\frac12\right] \)

Explanation:
1. Each probability lies in \( [0,1] \): \( -\frac13\le x\le\frac23 \), \( -3\le x\le1 \) and \( -\frac12\le x\le\frac12 \).
2. Mutual exclusivity gives \( P(A)+P(B)+P(C)\le1\Rightarrow\frac{3x+1}3+\frac{1-x}4+\frac{1-2x}2\le1\Rightarrow13-3x\le12\Rightarrow x\ge\frac13 \).
3. Combining, \( \frac13\le x\le\frac12 \). Answer: (a)

Teacher's Note:
1. Bound each probability.
2. Use the sum of probabilities \( \le1 \).

Answer: (a) \( \frac1{32} \)

Explanation:
1. \( npq=2 \) and \( np=4 \) give \( q=\frac12 \), \( p=\frac12 \) and \( n=8 \).
2. \( P(X=1)={}^8C_1\left(\frac12\right)^8=\frac8{256}=\frac1{32} \). Answer: (a)

Teacher's Note:
1. Find \( n \) and \( p \).

Answer: (a) \( \frac34 \)

Explanation:
1. \( P(\text{not solved})=\frac12\cdot\frac23\cdot\frac34=\frac14 \).
2. \( P(\text{solved})=1-\frac14=\frac34 \). Answer: (a)

Teacher's Note:
1. Use the complement.

Answer: (b) 65

Explanation:
1. Total marks of the class \( =100\times72=7200 \); boys' total \( =70\times75=5250 \).
2. Girls' total \( =1950 \) for 30 girls, so the average is \( 65 \). Answer: (b)

Teacher's Note:
1. Find the girls' total marks.

Answer: (b) \( \frac{24}{25} \)

Explanation:
1. Total outcomes \( =25\times25=625 \) and matching outcomes \( =25 \), so \( P(\text{prize})=\frac1{25} \).
2. \( P(\text{no prize})=1-\frac1{25}=\frac{24}{25} \). Answer: (b)

Teacher's Note:
1. Count matching pairs.

Answer: (d) None of these

Explanation:
1. Mutual exclusivity gives \( A\subseteq\overline B \), so \( P(A)\le P(\overline B) \), and the inequality need not be strict.
2. When \( A=\overline B \) we get equality, so (a) can fail; (b) and (c) can fail as well. Answer: (d)

Teacher's Note:
1. Use \( A\subseteq\overline B \).
2. Look for an equality case.

Answer: (b) \( \frac14 \)

Explanation:
1. Exactly one win in the first two matches and a win in the third.
2. \( P=2\cdot\frac12\cdot\frac12\cdot\frac12=\frac14 \). Answer: (b)

Teacher's Note:
1. List the orderings WLW and LWW.

Answer: (a) \( \frac13 \)

Explanation:
1. \( \frac25=qp+q^3p+q^5p+\cdots=\frac{qp}{1-q^2} \) with \( q=1-p \).
2. \( \frac{(1-p)p}{1-(1-p)^2}=\frac{1-p}{2-p}=\frac25\Rightarrow5-5p=4-2p \).
3. \( p=\frac13 \). Answer: (a)

Teacher's Note:
1. Sum the geometric series over even numbers of tosses.

Answer: (b) \( \dfrac{{}^7C_2\times5^5}{6^8} \)

Explanation:
1. Exactly two sixes in the first seven throws and a six on the eighth.
2. \( P={}^7C_2\left(\frac16\right)^2\left(\frac56\right)^5\cdot\frac16=\frac{{}^7C_2\times5^5}{6^8} \). Answer: (b)

Teacher's Note:
1. Fix the eighth throw as a six.
2. Use the binomial formula for the first seven.