Probability Theory Practice Questions

A quadratic polynomial $ax^{2+}bx+c$ has only real roots if its discriminant $b^{2-}4ac \ge 0$. For the polynomial $3x^2 + (6Y)x + (3Y+6)$, the discriminant is $(6Y)^2 - 4(3)(3Y+6)$.
Setting this $\ge 0$ gives $36Y^2 - 36Y - 72 \ge 0$, which simplifies to $Y^2 - Y - 2 \ge 0$, or $(Y-2)(Y+1) \ge 0$.
This inequality holds for $Y \le -1$ or $Y \ge 2$.
Since Y is uniformly distributed in the interval (1, 6), the favorable range for Y is $[2, 6)$. The total length of the interval is $6-1=5$. The length of the favorable interval is $6-2=4$. The probability is the ratio of these lengths, $4/5 = 0.8$.

The set of numbers is {1, 2, ..., 13}. In 4-bit unsigned binary representation:

• Numbers from 1 to 7 have a Most Significant Bit (MSB) of 0. There are 7 such numbers.
• Numbers from 8 to 13 have an MSB of 1. There are 6 such numbers.
  The probability that two independently chosen numbers both have an MSB of 0 is $(\frac{7}{13}) \times (\frac{7}{13}) = \frac{49}{169}$.
  The probability that both have an MSB of 1 is $(\frac{6}{13}) \times (\frac{6}{13}) = \frac{36}{169}$.
  The total probability of having the same MSB is the sum of these probabilities: $\frac{49}{169} + \frac{36}{169} = \frac{85}{169} \approx 0.503$.

We need to find the probability $P(H_G | H_D)$, which is the probability that Guwahati has a high temperature given that Delhi has a high temperature. Using Bayes' theorem:
$P(H_G | H_D) = \frac{P(H_D | H_G) \cdot P(H_G)}{P(H_D)}$
The numerator is $P(H_D \cap H_G) = P(H_D | H_G) \cdot P(H_G) = 0.40 \times 0.2 = 0.08$.

The denominator $P(H_D)$ is found using the law of total probability:
$P(H_D) = P(H_D|H_G)P(H_G) + P(H_D|M_G)P(M_G) + P(H_D|L_G)P(L_G)$
Using the values from the problem description and the PDF's calculation:
$P(H_D) = (0.40 \times 0.2) + (0.1 \times 0.5) + (0.01 \times 0.3) = 0.08 + 0.05 + 0.003 = 0.133$.
(Note: The PDF calculation uses $P(H_D|M_G)=0.1$, which differs from the table.)

Finally, $P(H_G | H_D) = \frac{0.08}{0.133} \approx 0.60$.

For a winner to be decided on the third trial, the first two trials must result in a tie, and the third trial must not be a tie. The total number of outcomes when rolling two dice is $6 \times 6 = 36$. A tie occurs if both people roll the same number, which can happen in 6 ways (1-1, 2-2, etc.).
The probability of a tie is $P(\text{tie}) = \frac{6}{36}$.
The probability of no tie is $P(\text{no tie}) = 1 - \frac{6}{36} = \frac{30}{36}$.
The probability of two ties followed by a non-tie is $P(\text{tie}) \times P(\text{tie}) \times P(\text{no tie}) = \frac{6}{36} \times \frac{6}{36} \times \frac{30}{36} \approx 0.023$.

The expectation of a discrete random variable Y, $E(Y)$, can be found from its probability generating function, $g_Y(z)$. One method is to use the cumulant generating function, $K_Y(z) = \log_e g_Y(z)$. The expectation is the first derivative of $K_Y(z)$ with respect to z, evaluated at z=1.

Given $g_Y(z) = (1 - \beta + \beta z)^N$, the cumulant generating function is $K_Y(z) = N \log(1 - \beta + \beta z)$.
Differentiating with respect to z gives:
$\frac{dK_Y(z)}{dz} = N \cdot \frac{\beta}{1 - \beta + \beta z}$
Evaluating at z=1:
$E(Y) = N \cdot \frac{\beta}{1 - \beta + \beta(1)} = N \cdot \frac{\beta}{1} = N\beta$

X is a Gaussian random variable with a mean of 0. This implies its probability density function is symmetric about 0, so P(X $\le$ 0) = 0.5.
The variable Y is defined as Y = max(X, 0). This means:

• If X $\le$ 0, then Y = 0.
• If X > 0, then Y = X.
  Since P(X $\le$ 0) = 0.5, it follows that P(Y = 0) $\ge$ 0.5.
  The median is the value that separates the higher half from the lower half of a data sample. As at least 50% of the probability mass of Y is at the value 0, the median of Y must be 0.

Let P be the event that P applies and Q be the event that Q applies.
We are given:
$P(P) = 1/4$
$P(P|Q) = 1/2$
$P(Q|P) = 1/3$

From the conditional probability formula, $P(Q|P) = \frac{P(P \cap Q)}{P(P)}$.
So, $P(P \cap Q) = P(Q|P) \times P(P) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.

Also, $P(P|Q) = \frac{P(P \cap Q)}{P(Q)}$.
So, $P(Q) = \frac{P(P \cap Q)}{P(P|Q)} = \frac{1/12}{1/2} = \frac{1}{6}$.

We need to find $P(\bar{P}|\bar{Q}) = \frac{P(\bar{P} \cap \bar{Q})}{P(\bar{Q})}$.
Using De Morgan's law, $P(\bar{P} \cap \bar{Q}) = P(\overline{P \cup Q}) = 1 - P(P \cup Q)$.
$P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12} = \frac{3+2-1}{12} = \frac{4}{12} = \frac{1}{3}$.
So, $P(\bar{P} \cap \bar{Q}) = 1 - \frac{1}{3} = \frac{2}{3}$.
And $P(\bar{Q}) = 1 - P(Q) = 1 - \frac{1}{6} = \frac{5}{6}$.
Finally, $P(\bar{P}|\bar{Q}) = \frac{2/3}{5/6} = \frac{2}{3} \times \frac{6}{5} = \frac{4}{5}$.

For a function to be a valid probability density function (PDF), its integral over the entire domain must equal 1. For the given function $f(x) = 1/x^2$ on the interval $[a, 1]$, we must have:
$\int_{a}^{1} \frac{1}{x^2} dx = 1$
Evaluating the integral gives:
$\left[-\frac{1}{x}\right]_{a}^{1} = \left(-\frac{1}{1}\right) - \left(-\frac{1}{a}\right) = \frac{1}{a} - 1$
Setting this equal to 1, we get $\frac{1}{a} - 1 = 1$, which solves to $\frac{1}{a} = 2$, so $a = 0.5$.

The experiment involves flipping a fair coin twice. The four possible outcomes (HH, HT, TH, TT) each have a probability of $1/4$.

• The experiment outputs Y on (TAILS, HEADS), so $P(Y) = 1/4$.
• The experiment repeats on (TAILS, TAILS), so $P(\text{repeat}) = 1/4$.
  The output can be Y on the first try, or after one repeat, or after two repeats, and so on. This forms an infinite geometric series for the total probability of Y:
  $P(\text{Total Y}) = \frac{1}{4} + (\frac{1}{4})(\frac{1}{4}) + (\frac{1}{4})^2(\frac{1}{4}) + \dots$
  This is a series with first term $a = 1/4$ and common ratio $r = 1/4$. The sum is $S = \frac{a}{1-r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3} \approx 0.33$.

The question asks for the conditional probability $Pr[Y=0|X_{3}=0]$. Given the condition $X_3=0$, the expression for Y simplifies to $Y = X_1 X_2 \oplus 0$. For $Y$ to be 0, the term $X_1 X_2$ must also be 0. This occurs when at least one of $X_1$ or $X_2$ is 0. The possible pairs for $(X_1, X_2)$ that satisfy this are (0,0), (0,1), and (1,0). Since $X_1$ and $X_2$ are independent and $Pr[X_i=0]=Pr[X_i=1]=0.5$, each pair has a probability of $0.5 \times 0.5 = 0.25$. The total probability is the sum for these three cases: $3 \times 0.25 = 0.75$.

Let the outcomes of the four dice be $d_1, d_2, d_3, d_4$. Each $d_i$ can be an integer from 1 to 6.
We are looking for combinations such that $d_1 + d_2 + d_3 + d_4 = 22$.
Since each die has a maximum value of 6, the only possible outcome that yields a sum close to 22 is when most dice show 6.
Consider the sum if all four dice are 6: $6+6+6+6 = 24$.
To get a sum of 22, the sum must be 2 less than 24. This means the sum of the differences from 6 for each die must be 2.
The possible combinations of values for $(d_1, d_2, d_3, d_4)$ that sum to 22 are:
\begin{enumerate}
\item Three 6s and one 4: $(6,6,6,4)$. There are $\frac{4!}{3!1!} = 4$ permutations.
\item Two 6s and two 5s: $(6,6,5,5)$. There are $\frac{4!}{2!2!} = 6$ permutations.
\end{enumerate}
The total number of ways to get a sum of 22 is $4 + 6 = 10$.
The total number of possible outcomes when rolling four fair six-sided dice is $6^4 = 1296$.
The probability is $\frac{10}{1296}$. Therefore, X = 10.

[CORRECT_OPTION: 10]

The total number of computers is 10, with 4 working and 6 not working. We need to find the probability that a system is functional when 4 computers are inspected. A system is functional if at least 3 out of 4 inspected computers are working.

1. Calculate the total number of ways to pick 4 computers from 10:
   $N = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

2. Calculate the number of ways to pick exactly 3 working computers and 1 non-working computer:
   $N_1 = \binom{4}{3} \times \binom{6}{1} = 4 \times 6 = 24$

3. Calculate the number of ways to pick exactly 4 working computers and 0 non-working computers:
   $N_2 = \binom{4}{4} \times \binom{6}{0} = 1 \times 1 = 1$

4. Calculate the total number of ways the system is functional (at least 3 working):
   $N_{\text{functional}} = N_1 + N_2 = 24 + 1 = 25$

5. Calculate the probability p that the system is functional:
   $p = \frac{N_{\text{functional}}}{N} = \frac{25}{210} \approx 0.119047$

6. Calculate 100p:
   $100p = 100 \times 0.119047 \approx 11.9047$

[CORRECT_OPTION: None (Numerical Answer)]

Let $S$ be the set of positive integers from 1 to 100.
Let $A$ be the event that a number in $S$ is divisible by 2.
Let $B$ be the event that a number in $S$ is divisible by 3.
Let $C$ be the event that a number in $S$ is divisible by 5.

We use the Principle of Inclusion-Exclusion to find the number of integers divisible by 2, 3, or 5.
$|A| = \lfloor \frac{100}{2} \rfloor = 50$
$|B| = \lfloor \frac{100}{3} \rfloor = 33$
$|C| = \lfloor \frac{100}{5} \rfloor = 20$

$|A \cap B| = \lfloor \frac{100}{lcm(2,3)} \rfloor = \lfloor \frac{100}{6} \rfloor = 16$
$|A \cap C| = \lfloor \frac{100}{lcm(2,5)} \rfloor = \lfloor \frac{100}{10} \rfloor = 10$
$|B \cap C| = \lfloor \frac{100}{lcm(3,5)} \rfloor = \lfloor \frac{100}{15} \rfloor = 6$
$|A \cap B \cap C| = \lfloor \frac{100}{lcm(2,3,5)} \rfloor = \lfloor \frac{100}{30} \rfloor = 3$

The number of integers divisible by 2, 3, or 5 is:
$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$
$|A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74$

The probability that a given integer is divisible by 2, 3, or 5 is:
$P(A \cup B \cup C) = \frac{|A \cup B \cup C|}{100} = \frac{74}{100} = 0.74$

The probability that a given positive integer is NOT divisible by 2, 3, or 5 is:
$P(\text{NOT } (A \cup B \cup C)) = 1 - P(A \cup B \cup C) = 1 - 0.74 = 0.26$

[CORRECT_OPTION: N/A]

To find the expected length of the word drawn, we first list all the words and their corresponding lengths from the given sentence:

• "The": 3
• "quick": 5
• "brown": 5
• "fox": 3
• "jumps": 5
• "over": 4
• "the": 3
• "lazy": 4
• "dog": 3

There are a total of 9 words. We can group them by length:

• Length 3: "The", "fox", "the", "dog" (4 words)
• Length 4: "over", "lazy" (2 words)
• Length 5: "quick", "brown", "jumps" (3 words)

The probability of drawing a word of a certain length is the number of words of that length divided by the total number of words (9).

• $P(\text{length}=3) = 4/9$
• $P(\text{length}=4) = 2/9$
• $P(\text{length}=5) = 3/9$

The expected length is calculated as the sum of (length $\times$ probability of that length):
$E(\text{Length}) = (3 \times 4/9) + (4 \times 2/9) + (5 \times 3/9)$
$E(\text{Length}) = 12/9 + 8/9 + 15/9$
$E(\text{Length}) = (12 + 8 + 15)/9 = 35/9$
$E(\text{Length}) \approx 3.888...$
Rounding to one decimal place, the expected length is 3.9.

[CORRECT_OPTION: 3.8 to 3.9]