Concept and Intuition

A discrete random variable's full probability distribution must sum to 1, and its mean is the weighted sum $\sum x_iP(X{=}x_i)$. With two of the four probabilities related ($P(X{=}3)=2P(X{=}1)$) and one given numerically ($P(X{=}2)=0.3$), we get exactly two equations for the two remaining unknowns.

Step-by-Step Solution

1. Let $P(X{=}0)=p_0$, $P(X{=}1)=p_1$. Then $P(X{=}2)=0.3$, $P(X{=}3)=2p_1$.
2. Normalization: $p_0+p_1+0.3+2p_1=1\Rightarrow p_0+3p_1=0.7$.
3. Mean: $0\cdot p_0+1\cdot p_1+2(0.3)+3(2p_1)=1.3\Rightarrow p_1+0.6+6p_1=1.3\Rightarrow7p_1=0.7\Rightarrow p_1=0.1$.
4. Substitute back: $p_0=0.7-3(0.1)=0.7-0.3=0.4=\dfrac25$.

Common Mistakes

• Forgetting that the mean equation also needs the $P(X{=}2)$ term included with weight $2$.
• Solving for $p_0$ before finding $p_1$ unnecessarily — solve the mean equation for $p_1$ first.

ANSWER: B

Concept and Intuition

A probability distribution's values must sum to exactly 1. This gives an equation in $k$ that we solve, keeping only the root that yields non-negative probabilities. Then $P(X\le2)$ is simply the sum of the probabilities at $x=0,1,2$.

Step-by-Step Solution

1. Sum all probabilities and set equal to 1:

2. Solve the quadratic using the formula:

So $k=\dfrac{2}{20}=\dfrac1{10}$ or $k=\dfrac{-20}{20}=-1$.

3. Reject $k=-1$ (gives negative probabilities); take $k=0.1$.

4. Verify all probabilities are valid: $P(0)=0.3,\ P(1)=0.5,\ P(2)=0.03,\ P(3)=0.14,\ P(4)=0.03$; sum $=1.0$. ✓

5. Compute:

Common Mistakes

• Accepting the extraneous negative root $k=-1$.
• Forgetting to include $P(X=2)=3k^2$ (only summing $P(0)+P(1)$).

ANSWER: D

Concept and Intuition

A probability distribution assigns a probability $P(X=x)$ to every possible outcome, and since the outcomes are exhaustive and mutually exclusive, the total probability must be exactly $1$. This single normalization equation is usually enough to pin down an unknown constant like $k$.

Step-by-Step Solution

1. List the probabilities: $P(X=1)=2k,\ P(X=2)=4k,\ P(X=3)=3k,\ P(X=4)=k$.
2. Sum them and set equal to $1$: $2k+4k+3k+k = 10k = 1$.
3. Solve: $k = \dfrac{1}{10}$.
4. Sanity check: each probability ($0.2, 0.4, 0.3, 0.1$) is between $0$ and $1$, so this is a valid distribution.

Common Mistakes

• Forgetting to include all four terms in the sum.
• Not checking that the resulting probabilities are all non-negative and $\le 1$.

ANSWER: A

Concept and Intuition

For any discrete random variable, the probabilities over its whole range must sum to exactly 1. Writing that sum-to-1 condition down turns the problem into solving a polynomial equation in $C$ — but since $C$ is a probability parameter, we must also sanity-check that the resulting individual probabilities are all non-negative and at most 1 (a root of the polynomial that fails this check would be spurious/extraneous for a genuine probability distribution).

Step-by-Step Solution

1. Sum-to-one condition: $P(X=0)+P(X=1)+P(X=2)=1$ $3C^3+(4C-10C^2)+(5C-1)=1$.
2. Simplify: $3C^3-10C^2+9C-1=1 \Rightarrow 3C^3-10C^2+9C-2=0$.
3. Try the given options as roots. $C=\frac13$: $3\cdot\frac{1}{27}-10\cdot\frac19+9\cdot\frac13-2 = \frac{1}{9}-\frac{10}{9}+3-2=-1+1=0$. This works.
4. Verify the resulting probabilities are valid: $P(X=0)=\frac19$, $P(X=1)=\frac43-\frac{10}{9}=\frac{12-10}{9}=\frac29$, $P(X=2)=\frac53-1=\frac23$. Sum $=\frac19+\frac29+\frac69=1$ ✓, and each lies in $[0,1]$.
5. So $C=\frac13$ is the consistent answer.

Common Mistakes

• Accepting a root of the cubic without checking that each individual probability stays within $[0,1]$ (a cubic can have roots that make some $P(X=i)$ negative or exceed 1, which must be rejected).
• Arithmetic slips combining the $-10C^2$ and $+9C$ terms when simplifying the sum-to-one equation.

ANSWER: B

Concept and Intuition

When probabilities of mutually exclusive, exhaustive events are given in a chain of ratios, express everything as a multiple of the smallest and normalize using the fact that all probabilities must add to 1.

Step-by-Step Solution

1. Let $P(C) = x$.
2. $P(B) = 2\,P(C) = 2x$.
3. $P(A) = 2\,P(B) = 4x$.
4. Since exactly one of A, B, C wins, $P(A)+P(B)+P(C) = 1 \Rightarrow 4x+2x+x = 7x = 1 \Rightarrow x = \dfrac{1}{7}$.
5. So $P(A) = \dfrac{4}{7}$, $P(B) = \dfrac{2}{7}$, $P(C) = \dfrac{1}{7}$.

Common Mistakes

• Reversing the ratio (thinking $P(A)$ is half of $P(B)$ instead of twice).
• Forgetting that the three probabilities must sum to 1 exactly.

ANSWER: A

Concept and Intuition

Probability is monotonic with respect to subset inclusion: a smaller (more restrictive) event can never be more likely than a larger event that contains it. If $E_1\subseteq E_2$, every outcome favouring $E_1$ also favours $E_2$, so $P(E_1)\le P(E_2)$.

Step-by-Step Solution

1. (i) $P(\phi)=0$ — a standard axiom, true.
2. (ii) $P(E^c)=1-P(E)$ — follows since $E$ and $E^c$ partition $S$, true.
3. (iii) $0\le P(E)\le1$ for every $E\subseteq S$ — the basic probability axiom, true.
4. (iv) claims "$E_1\subseteq E_2 \Rightarrow P(E_2)\le P(E_1)$" — this has the inequality backwards; the correct statement is $P(E_1)\le P(E_2)$. So (iv) is false in general, hence not satisfied by $P$.

Common Mistakes

• Misreading the direction of the inclusion/inequality and picking (iii) or (ii) instead, which are both correctly stated axioms.

ANSWER: B

Concept and Intuition

With two fair six-sided dice, there are $6\times6=36$ equally likely ordered outcomes. Each possible total from 2 to 12 has a known number of ways it can occur (a classic triangular-then-decreasing pattern: 1,2,3,4,5,6,5,4,3,2,1 for sums 2 through 12). We simply sum the counts corresponding to prime totals.

Step-by-Step Solution

1. Total outcomes when rolling two dice: $6\times6=36$.
2. Number of ways to get each sum: sum=2→1 way, 3→2, 4→3, 5→4, 6→5, 7→6, 8→5, 9→4, 10→3, 11→2, 12→1.
3. Identify which sums from 2 to 12 are prime: 2, 3, 5, 7, 11 (4, 6, 8, 9, 10, 12 are composite; 1 is not a possible sum here anyway).
4. Sum the corresponding counts: for sum 2: 1 way; sum 3: 2 ways; sum 5: 4 ways; sum 7: 6 ways; sum 11: 2 ways. Total favorable $= 1+2+4+6+2 = 15$.
5. Probability $=\dfrac{15}{36} = \dfrac{5}{12}$ (dividing numerator and denominator by 3).

Common Mistakes

• Forgetting that 1 is not prime, and mistakenly not including 2 as prime (2 IS prime, and it's the smallest possible sum, achieved by (1,1)).
• Missing one of the prime sums (like 11) or mis-remembering the count-per-sum table.
• Forgetting to reduce the fraction $15/36$ to lowest terms.

ANSWER: B

Concept and Intuition

Any valid probability mass function must have its probabilities sum to exactly 1 over its whole range. Recognizing $\sum_{x=0}^\infty \frac{C^x}{x!}=e^C$ as the Taylor series of $e^C$ lets us convert the normalization condition into a simple exponential equation.

Step-by-Step Solution

1. Since $X$ takes values $1,2,3,\dots$, normalization requires $\displaystyle\sum_{x=1}^{\infty} P(X=x) = 1$, i.e. $\displaystyle\sum_{x=1}^{\infty}\frac{C^x}{x!}=1$.
2. Recall the Taylor series $e^C = \displaystyle\sum_{x=0}^{\infty}\frac{C^x}{x!} = 1 + \sum_{x=1}^{\infty}\frac{C^x}{x!}$.
3. So $\displaystyle\sum_{x=1}^{\infty}\frac{C^x}{x!} = e^C - 1$.
4. Set this equal to 1: $e^C - 1 = 1 \Rightarrow e^C = 2$.
5. Taking natural log of both sides: $C = \log 2$.
6. Check validity: $\log 2 \approx 0.693 > 0$, so each $P(X=x)=\frac{C^x}{x!}>0$ as required for a probability.

Common Mistakes

• Forgetting to subtract the $x=0$ term ($=1$) when relating the sum starting at $x=1$ to the full exponential series, which would incorrectly give $C=\ln1=0$.

ANSWER: C

Concept and Intuition

For probabilities to make sense, $0\le P(A),P(B),P(C)\le 1$. Mutual exclusivity means $P(A\cup B\cup C)=P(A)+P(B)+P(C)\le 1$ (it could be less than 1 if there's a residual sample-space probability outside all three). All these inequalities on $x$ must hold simultaneously.

Step-by-Step Solution

1. $0\le\dfrac{3x+1}{3}\le1 \Rightarrow -1\le 3x\le 2 \Rightarrow -\dfrac13\le x\le \dfrac23$.
2. $0\le\dfrac{1-x}{4}\le1 \Rightarrow -3\le x\le 1$.
3. $0\le\dfrac{1-2x}{2}\le1 \Rightarrow -\dfrac12\le x\le \dfrac12$.
4. Sum condition: $\dfrac{3x+1}{3}+\dfrac{1-x}{4}+\dfrac{1-2x}{2}\le 1$. Over a common denominator 12: $\dfrac{4(3x+1)+3(1-x)+6(1-2x)}{12}\le1 \Rightarrow \dfrac{-3x+13}{12}\le1 \Rightarrow -3x\le-1 \Rightarrow x\ge\dfrac13$.
5. Intersecting all four: $x\ge\dfrac13$ (from step 4), $x\le\dfrac12$ (from step 3, the tightest upper bound), and the other bounds are looser. So $x\in\left[\dfrac13,\dfrac12\right]$.

Common Mistakes

• Only checking the individual $[0,1]$ bounds and forgetting the sum-at-most-1 condition (or vice versa) — both are needed to pin down the interval.
• Arithmetic slips combining the three fractions over a common denominator.

ANSWER: A

Concept and Intuition

$x^2+bx+c=0$ has real roots exactly when the discriminant $b^2-4c\geq0$, i.e., $c \le b^2/4$. Since $b,c$ are each chosen uniformly from $\{1,2,3,4,5,6\}$, there are $6\times6=36$ equally likely pairs total.

Step-by-Step Solution

1. For $b=1$: need $c\le0.25$ — no valid $c$ (0 pairs).
2. For $b=2$: need $c\le1$ — $c=1$ (1 pair).
3. For $b=3$: need $c\le2.25$ — $c=1,2$ (2 pairs).
4. For $b=4$: need $c\le4$ — $c=1,2,3,4$ (4 pairs).
5. For $b=5$: need $c\le6.25$ — all $c=1,\ldots,6$ qualify (6 pairs).
6. For $b=6$: need $c\le9$ — all $c=1,\ldots,6$ qualify (6 pairs).
7. Total favorable pairs $=0+1+2+4+6+6=19$.
8. Probability $=\dfrac{19}{36}$.

Common Mistakes

• Forgetting the discriminant must be $\geq0$ (not $>0$), which would wrongly exclude the boundary case $b^2=4c$.

ANSWER: C