Probability — JEE MCQs with answers

Answer: D — Let $E_1$: the ball is drawn from box I; $E_2$: the ball is drawn from box II; $E_3$: the ball is drawn from box III; and $A$: the ball drawn is red. Priors: $P(E_1) = \dfrac{1}{6}$, $P(E_2) = \dfrac{1}{3}$, $P(E_3) = \dfrac{1}{2}$. Likelihoods: $P(A|E_1) = \dfrac{3}{5}$, $P(A|E_2) = \dfrac{1}{5}$, $P(A|E_3) = \dfrac{2}{5}$. The products of prior and likelihood, in the same order, are $\dfrac{1}{10}$, $\dfrac{1}{15}$, $\dfrac{1}{5}$; their sum is $P(A) = \dfrac{11}{30}$. By Bayes' theorem, $P(E_1|A) = \dfrac{1}{10} \div \dfrac{11}{30} = \dfrac{3}{11}$. So the answer is option D. Option A ($\dfrac{1}{5}$) makes both of these slips: equal priors $\dfrac{1}{3}$, and no division by $P(A)$. Option B ($\dfrac{1}{2}$) takes the three boxes as equally likely, although the die gives them probabilities $\dfrac{1}{6}$, $\dfrac{1}{3}$ and $\dfrac{1}{2}$. Option C ($\dfrac{1}{10}$) is the product $\dfrac{1}{6} \times \dfrac{3}{5}$ left undivided by $P(A)$.

Answer: A — The step is the rule for the probability of a union, $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$, applied to $X = A \cap F$ and $Y = B \cap F$; their intersection is $X \cap Y = A \cap B \cap F$, which is the term subtracted (p.409). So the answer is option A. $P(X \cup Y) = P(X) + P(Y)$ would need $X$ and $Y$ to be disjoint and would leave no third term; the book uses it only afterwards, for disjoint $A$ and $B$. $P(X \cap Y) = P(X) \cdot P(Y)$ and $P(X') = 1 - P(X)$ are not used anywhere in this proof. Dividing the three terms by $P(F)$ then gives $P(A|F) + P(B|F) - P((A \cap B)|F)$.

Answer: C — The event happens in two ways that cannot occur together. First a red ball, then a blue ball: $\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{4}{15}$ (after a red ball is removed, 4 of the 9 balls left are of the other kind). First a blue ball, then a red ball: $\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{4}{15}$. Adding the two products, the probability is $\dfrac{4}{15} + \dfrac{4}{15} = \dfrac{8}{15}$. Counting agrees: of the $10 \times 9 = 90$ equally likely ordered pairs of balls, $2 \times 6 \times 4 = 48$ are of this type. This is option C. Option D counts only one of the two orders: $\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{4}{15}$, which is the probability of 'first a red ball, then a blue ball'. Option A counts both orders but keeps 10 as the total for the second draw, as if the first ball were put back: $2 \times \dfrac{6}{10} \times \dfrac{4}{10} = \dfrac{12}{25}$. Option B makes both of these slips: $\dfrac{6}{10} \times \dfrac{4}{10} = \dfrac{6}{25}$.

Answer: A — Remark (iii) reads: two experiments are said to be independent if for every pair of events E and F, where E is associated with the first experiment and F with the second experiment, the probability of the simultaneous occurrence of the events E and F when the two experiments are performed is the product of P(E) and P(F) calculated separately on the basis of two experiments. The relation is required of every such pair. This is option A. Option B weakens 'every pair' to one pair. Options C and D add restrictions (equal probabilities, mutually exclusive) that the Remark does not make.

Answer: B — A leap year has 366 days, that is 52 complete weeks and 2 more days ($366 = 52 \times 7 + 2$). The 52 weeks give 52 Tuesdays, so the year has 53 Tuesdays exactly when one of the 2 remaining days is a Tuesday. These 2 days are consecutive, and the first of them is equally likely to be any of the 7 days of the week, giving 7 equally likely pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Two of them, (Mon, Tue) and (Tue, Wed), contain a Tuesday, so the probability is $\dfrac{2}{7}$, which is option B and is the answer printed for Question 5 of the Miscellaneous Exercise. Option A takes the year to have 365 days (one remaining day instead of two); this gives $\dfrac{1}{7}$. Option C is the probability of the complementary event, that the year has only 52 Tuesdays: $1 - \dfrac{2}{7} = \dfrac{5}{7}$. Option D makes both slips: $1 - \dfrac{1}{7} = \dfrac{6}{7}$.

Answer: C — $P(E) = \dfrac{1}{2}$. Testing $P(E \cap F) = P(E)\,P(F)$ for each event: for $F_1$: $P(F_1) = \dfrac{1}{2}$, $P(E \cap F_1) = \dfrac{1}{4}$, $P(E)\,P(F_1) = \dfrac{1}{4}$, so independent; for $F_2$: $P(F_2) = \dfrac{1}{2}$, $P(E \cap F_2) = \dfrac{1}{4}$, $P(E)\,P(F_2) = \dfrac{1}{4}$, so independent; for $F_3$: $P(F_3) = \dfrac{1}{6}$, $P(E \cap F_3) = \dfrac{1}{12}$, $P(E)\,P(F_3) = \dfrac{1}{12}$, so independent; for $F_4$: $P(F_4) = \dfrac{1}{4}$, $P(E \cap F_4) = 0$, $P(E)\,P(F_4) = \dfrac{1}{8}$, so not independent. Hence E is independent of exactly 3 of the four events. This is option C. Option A: it counts only $F_4$, holding that events with no common outcome are independent and that events with a common outcome are dependent. Option B: it compares $P(E|F)$ with $P(F)$ instead of with $P(E)$; that agrees for $F_1$ and $F_2$ only, because $P(E|F_3) = \dfrac{1}{2}$ is compared with $P(F_3) = \dfrac{1}{6}$. Option D: it also counts $F_4$, holding that events which cannot occur together are independent.

Answer: B — $F = \{\mathrm{HHT},\ \mathrm{HTT},\ \mathrm{THT},\ \mathrm{TTT}\}$, so $n(F) = 4$, and $E \cap F = \{\mathrm{HTT},\ \mathrm{THT}\}$ is listed correctly ($\mathrm{TTH}$ has exactly one head but its third toss is a head, so it is not in F). The conditional probability is $\dfrac{n(E \cap F)}{n(F)} = \dfrac{2}{4} = \dfrac{1}{2}$: the student divided by $n(S) = 8$ instead of $n(F) = 4$. This is option B. Option A divides by $n(E)$, which gives $P(F|E)$, not $P(E|F)$. Option C takes $E \cap F$ to be the whole of E and keeps the wrong denominator $n(S)$. Option D corrects the denominator but takes $E \cap F$ to be the whole of E, although $\mathrm{TTH}$ is not in F.

Answer: A — Let $E$ be the event that the e-mail is spam, $E'$ the event that it is not, and $A$ the event that the filter marks it as spam, so that $A'$ is the event that it is not marked. Then $P(E) = 0.2$, $P(E') = 0.8$, $P(A' \mid E) = 1 - 0.95 = 0.05$ and $P(A' \mid E') = 1 - 0.05 = 0.95$. By Bayes' theorem, $P(E \mid A') = \dfrac{P(E)P(A' \mid E)}{P(E)P(A' \mid E) + P(E')P(A' \mid E')} = \dfrac{0.2 \times 0.05}{0.2 \times 0.05 + 0.8 \times 0.95} = \dfrac{0.01}{0.77} = \dfrac{1}{77}$. This is option A. Option B leaves out the priors and also does not divide by $P(A')$: $\dfrac{1}{2} \times P(A' \mid E) = \dfrac{1}{40}$. Option C stops at the numerator $P(E)\,P(A' \mid E) = \dfrac{1}{100}$ and does not divide by $P(A')$. Option D leaves out the priors (treats the two hypotheses as equally likely): $\dfrac{P(A' \mid E)}{P(A' \mid E) + P(A' \mid E')} = \dfrac{1}{20}$.

Answer: C — $P(\text{first good}) = \dfrac{9}{12}$, $P(\text{second good} \mid \text{first good}) = \dfrac{8}{11}$, and with two good pens removed all 3 defective pens are among the 10 left: $\dfrac{3}{10}$. Product: $\dfrac{9}{12} \times \dfrac{8}{11} \times \dfrac{3}{10}$ $=$ $\dfrac{9}{55}$. This is option C. Option A: multiplies by 3 for the position of the defective pen, although the stem fixes it as the third; this gives $\dfrac{27}{55}$. Option B: keeps 12 pens in the box for all three draws while still reducing the number of good pens (factors $\dfrac{9}{12}$, $\dfrac{8}{12}$, $\dfrac{3}{12}$); this gives $\dfrac{1}{8}$. Option D: puts each pen back before the next is drawn (with replacement), so that the factors are $\dfrac{9}{12}$, $\dfrac{9}{12}$, $\dfrac{3}{12}$; this gives $\dfrac{9}{64}$.

Answer: A — p.406: 'We have also established equivalence between the axiomatic theory and the classical theory of probability in case of equally likely outcomes.' The classical theory counts favourable outcomes against all outcomes, which is meaningful only when the outcomes are equally likely, so the answer is option A. Options B and C name relations between two events (mutually exclusive, independent), not a condition on the outcomes of the experiment, and the Introduction attaches neither to the equivalence. Option D drops the condition altogether: when the outcomes are not equally likely, counting outcomes does not give the probability.