Probability — JEE MCQs with answers

Answer: C — Option C is correct: the three events are $\{HHH\}$, $\{HHT, HTH, THH\}$ and $\{HTT, THT, TTH, TTT\}$, holding $1+3+4=8$ outcomes with no repetition, so they are pairwise disjoint and their union is S. Option A is wrong: it stops at exactly two tails, so TTT lies in none of the three events. Option B is wrong: the first two events already cover S, so the third overlaps both of them; HHT, for instance, lies in the first and the third. Option D is wrong: 'exactly two heads appear' is contained inside 'at least two heads appear', so those two events share three outcomes.

Answer: B — The heading of the section is 'Probability of event ‘not A’', and it works out $P(A)=\frac{2}{5}$ and $P(A')=\frac{3}{5}$ for that ten card example before proving $P(A')=1-P(A)$. Option A is the title of the whole of §14.2, not of this section. Option C names the section that proves the addition rule. Option D names the section that gives $P(E)=\frac{m}{n}$.

Answer: C — Option C is correct: $B=\{(5, 6), (6, 5)\}$, while every outcome of A has 2 as its first entry. No outcome of B has first entry 2, so $A\cap B=\phi$ and the two events can not occur simultaneously. Option A is wrong: $(5, 6)$ has first entry 5, so it lies in B only. Option B is wrong: A has 6 outcomes and B has 2, so $A\cup B$ has only 8 of the 36 outcomes and is far from S. Option D is wrong: $(6, 5)$ has first entry 6, so it too lies in B only.

Answer: A — Option A is correct: If $\omega_3$ is left out, $P(E)=0.05+0.15+0.5=0.7$. Leaving out any other single outcome gives 0.5, 0.85 or 0.95. Option B is wrong: leaving out $\omega_1$ gives $P(E)=0.15+0.3+0.5=0.95$. Option C is wrong: leaving out $\omega_2$ gives $P(E)=0.05+0.3+0.5=0.85$. Option D is wrong: leaving out $\omega_4$ gives $P(E)=0.05+0.15+0.3=0.5$.

Answer: A — Option A is correct: only TTT carries no head, so F holds the remaining seven outcomes and has more than one sample point. Option B is wrong: that count keeps TTT as well and gives the whole sample space. Option C is wrong: three is the count for 'exactly one head appeared'. Option D is wrong: a simple event has only one sample point.

Answer: C — C is correct: Treat OA as one block: the block and M, N, D, Y give $5!=120$ orders, and the block can be OA or AO, so $n(E)=2\times 120=240$. With $n(S)=6!=720$, $P=\frac{240}{720}=\frac{1}{3}$. A is wrong: Arranging the block and the other letters in $4!$ ways instead of $5!$ gives $\frac{1}{15}$. B is wrong: Treating the block as OA only, forgetting AO gives $\frac{1}{6}$. D is wrong: Moving the block through its 5 positions but not rearranging M, N, D, Y gives $\frac{1}{72}$.

Answer: A — Option A is correct: two heads and one tail can fall in three orders, HHT, HTH and THH. Option B is wrong: it reads the description as 'at least two heads' and adds HHH. Option C is wrong: it lists the outcomes with two tails and one head. Option D is wrong: it records only one of the three orders, as though the order of the tosses did not matter.

Answer: D — $P(A\cup B\cup C)=0.4+0.3+0.35-0.12-0.15-0.1+0.05=1.05-0.37+0.05=0.73$. Option A subtracts $P(A\cap B\cap C)$ instead of adding it. Option B subtracts only $P(A\cap B\cap C)$ from the sum of the three single probabilities. Option C leaves out the final $+P(A\cap B\cap C)$.

Answer: C — Option C is correct: $A=\{TTT\}$ and $B=\{HTT, THT, TTH\}$ together hold the four outcomes with at most one head, so the four outcomes with two or more heads are what is missing, and that is exactly the event 'at least two heads appear'. Option A is wrong: $\{HHT, HTH, THH\}$ still leaves HHH out of the union. Option B is wrong: $\{HHH, HHT, HTH, HTT\}$ leaves THH out of the union. Option D is wrong: $\{HTT, THT, TTH, TTT\}$ is already contained in $A\cup B$, so it adds nothing and four outcomes stay uncovered.

Answer: B — Option B is correct: The four probabilities must add up to 1: $P(\omega_4)=1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=1-1=0$. A probability of 0 is allowed, just as the entries 0 are allowed in Example 4(b). Option A is wrong: it takes $\frac{1}{n}=\frac{1}{4}$ as if the outcomes were equally likely; then the four values would add up to $\frac{5}{4}$. Option C is wrong: it copies $P(S)=1$ for the remaining outcome, making the total 2. Option D is wrong: it copies $P(\omega_3)=\frac{1}{6}$, perhaps rejecting 0 as a probability; the total would then be $\frac{7}{6}$.