Probability Theory Practice Questions

The number of cars $p$ follows a Poisson distribution with mean $\lambda = 3$.
The probability mass function for a Poisson distribution is given by $P(p=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.
We need to find the probability of observing fewer than 3 cars, which means $P(p < 3)$.
This is equivalent to $P(p=0) + P(p=1) + P(p=2)$.
Substituting $\lambda = 3$ into the formula:
$P(p < 3) = \frac{3^0 e^{-3}}{0!} + \frac{3^1 e^{-3}}{1!} + \frac{3^2 e^{-3}}{2!}$
$P(p < 3) = e^{-3} + 3e^{-3} + \frac{9e^{-3}}{2} = e^{-3} \left(1 + 3 + \frac{9}{2}\right) = e^{-3} \left(\frac{2+6+9}{2}\right) = \frac{17e^{-3}}{2} = \frac{17}{2e^3}$

A random variable $X$ takes values $-1$ and $+1$ with equal probability $P(X=-1) = 0.5$ and $P(X=+1) = 0.5$.
The cumulative distribution function (CDF) $F(x)$ is defined as $F(x) = P(X \le x)$.
For $x = -1$: $F(-1) = P(X \le -1)$. Since $-1$ is the smallest value $X$ can take, this is simply $P(X=-1) = 0.5$.
For $x = +1$: $F(+1) = P(X \le +1)$. This includes all possible values of $X$, so $F(+1) = P(X=-1) + P(X=+1) = 0.5 + 0.5 = 1$.
Thus, the values of the CDF at $x=-1$ and $x=+1$ are $0.5$ and $1$, respectively.

Let $D_1$ be the outcome of the first roll and $D_2$ be the outcome of the second roll (if applicable).
The sample space for the first roll is $S_1 = \{1, 2, 3, 4, 5, 6\}$, with $P(D_1=k) = 1/6$ for each $k \in S_1$.

If $D_1 \in \{1, 2, 3\}$, the die is rolled a second time. The sample space for the second roll is $S_2 = \{1, 2, 3, 4, 5, 6\}$, with $P(D_2=k) = 1/6$ for each $k \in S_2$.
We need to find the probability that the sum $D_1+D_2 \ge 6$. The favorable outcomes for $D_2$ given $D_1$ are:

• If $D_1=1$, then $D_2 \in \{5, 6\}$ (sum $\ge 6$). $P(D_1=1 \text{ and } D_2 \in \{5,6\}) = (1/6) \times (2/6) = 2/36$.
• If $D_1=2$, then $D_2 \in \{4, 5, 6\}$ (sum $\ge 6$). $P(D_1=2 \text{ and } D_2 \in \{4,5,6\}) = (1/6) \times (3/6) = 3/36$.
• If $D_1=3$, then $D_2 \in \{3, 4, 5, 6\}$ (sum $\ge 6$). $P(D_1=3 \text{ and } D_2 \in \{3,4,5,6\}) = (1/6) \times (4/6) = 4/36$.

If $D_1 \in \{4, 5, 6\}$, the die is not rolled a second time. In this case, the sum is simply $D_1$. We need $D_1 \ge 6$.

• If $D_1=4$, sum is 4, which is not $\ge 6$.
• If $D_1=5$, sum is 5, which is not $\ge 6$.
• If $D_1=6$, sum is 6, which is $\ge 6$. $P(D_1=6) = 1/6 = 6/36$.

The total probability is the sum of probabilities of these favorable cases:
P(\text{\sum} \ge 6) = 2/36 + 3/36 + 4/36 + 6/36 = (2+3+4+6)/36 = 15/36 = 5/12.

Let S be the set of possible pairs of cards drawn, where order matters. The total number of ways to draw two cards one at a time from a deck of 5 is given by the permutation formula $P(n, k) = n! / (n-k)!$. Here, $n=5$ and $k=2$.
$P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$
Let E be the event that the number on the first card is one higher than the number on the second card. The favorable pairs (first card, second card) are:
(2, 1), (3, 2), (4, 3), (5, 4).
There are 4 such favorable outcomes.
The probability of event E is the ratio of favorable outcomes to the total number of possible outcomes:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = \frac{1}{5}$

Let $A$ be the event that both outcomes are heads (HH).
Let $B$ be the event that at least one outcome is a head (HT, TH, HH).
The sample space for flipping two fair coins is {HH, HT, TH, TT}.
Given that at least one outcome is a head, the reduced sample space for event $B$ is {HH, HT, TH}. There are 3 equally likely outcomes in $B$.
Out of these outcomes, only one (HH) corresponds to event $A$ (both outcomes are heads).
The probability of event $A$ given event $B$ is $P(A|B) = \frac{\text{Number of outcomes \in } (A \cap B)}{\text{Number of outcomes \in } B}$.
Since $A \cap B$ is {HH}, the number of outcomes in $A \cap B$ is 1.
Thus, $P(A|B) = \frac{1}{3}$.

Let $X$ be a random variable.
The variance of $X$ is defined as Var(X) = E\[X^2]$ - (E[X])^2$. The problem statement defines$R$as the difference between the expectation of the square of the random variable$(E$[X^2]$)$and the square of the expectation of the random variable$(E[X])^2$. So,$R = E$[X^2]$ - (E[X])^2$. By definition,$R = Var(X)$. The variance of any random variable is always non-negative. Therefore,$Var(X) \ge 0$, which implies$R \ge 0$.

Let $X = [x_1, x_2, ..., x_n]$ be a sequence of random values with mean $\mu_x$ and standard deviation $\sigma_x$.
A new sequence $Y$ is derived as $y_i = a \cdot x_i + b$, where $a$ and $b$ are positive constants.
The mean of $Y$ is given by $\mu_y = E[Y] = E[aX+b] = aE[X]+b = a\mu_x+b$. So, statement C is correct.
The standard deviation of $Y$ is given by $\sigma_y = SD[Y] = SD[aX+b]$. For a linear transformation, $SD[aX+b] = |a|SD[X] = a\sigma_x$ (since $a$ is positive). So, statement D, which says $\sigma_y = a\sigma_x + b$, is incorrect because the additive constant $b$ does not affect the standard deviation.
The mode represents the most frequent value, and the median represents the middle value. A linear transformation $y_i = a \cdot x_i + b$ (with $a>0$) preserves the relative order and frequency of data points. Therefore, the index position of the mode and median will remain the same. Thus, statements A and B are correct.

To solve this problem, we need to find the total number of divisors of $10^{99}$ and the number of those divisors that are multiples of $10^{96}$. The ratio of these two numbers will give us the required probability.

1. Total Number of Divisors (Total Outcomes):
   First, we find the prime factorization of the number: $10^{99} = (2 \times 5)^{99} = 2^{99} \times 5^{99}$.
   Any divisor of this number will be of the form $2^a \times 5^b$, where the exponents 'a' and 'b' can range from 0 to 99.
   The number of choices for 'a' is $99+1 = 100$.
   The number of choices for 'b' is $99+1 = 100$.
   The total number of divisors is the product of these choices: $100 \times 100 = 10000$.

2. Favorable Divisors (Favorable Outcomes):
   We are looking for divisors that are multiples of $10^{96}$. Let the divisor be $D$. For $D$ to be a multiple of $10^{96}$, the exponents in its prime factorization must be greater than or equal to the exponents in the prime factorization of $10^{96}$.
   The prime factorization of the multiple is $10^{96} = 2^{96} \times 5^{96}$.
   So, for our divisor $D = 2^a \times 5^b$, we must have $a \ge 96$ and $b \ge 96$.
   Since $D$ is a divisor of $10^{99}$, we also know $a \le 99$ and $b \le 99$.
   Thus, the possible values for 'a' are {96, 97, 98, 99}, which are 4 choices.
   Similarly, the possible values for 'b' are {96, 97, 98, 99}, which are 4 choices.
   The total number of favorable divisors is $4 \times 4 = 16$.

3. Calculate the Probability:
   Probability = (Favorable Outcomes) / (Total Outcomes) = $\frac{16}{10000}$.
   Simplifying this fraction gives $\frac{1}{625}$.

To find the probability of a computer being declared faulty, we can use the law of total probability. A computer can be declared faulty in two mutually exclusive scenarios:

1. The computer is actually faulty, and the test correctly identifies it as faulty.
2. The computer is not faulty, and the test incorrectly identifies it as faulty.

Let's define the events:

• $F$: The event that a computer has a faulty assembly. $P(F) = p$.
• $NF$: The event that a computer is not faulty. $P(NF) = 1 - p$.
• $DF$: The event that a computer is declared faulty by the test.

We are given that the test gives the correct result with probability $q$. This can be interpreted as:

• The probability of declaring a faulty computer as faulty is $q$. This is the conditional probability $P(DF | F) = q$.
• The probability of declaring a non-faulty computer as non-faulty is also $q$. Therefore, the probability of incorrectly declaring a non-faulty computer as faulty is $1-q$. This is $P(DF | NF) = 1 - q$.

The total probability of being declared faulty, $P(DF)$, is the sum of the probabilities of the two scenarios:

$P(DF) = P(\text{scenario 1}) + P(\text{scenario 2})$

Using the formula for conditional probability, $P(A \cap B) = P(A|B) \times P(B)$:

• Probability of Scenario 1 (is faulty AND declared faulty):
  $P(DF \cap F) = P(DF | F) \times P(F) = q \times p = pq$

• Probability of Scenario 2 (is not faulty AND declared faulty):
  $P(DF \cap NF) = P(DF | NF) \times P(NF) = (1 - q) \times (1 - p)$

Summing these gives the total probability:

$P(DF) = pq + (1 - p)(1 - q)$

Let $P(i)$ be the probability of face $i$. Let $O$ be odd faces, $E$ be even faces.
$P(O) = 0.9 P(E)$ and $P(O) + P(E) = 1 \Rightarrow P(E) = 1/(1+0.9) = 10/19$.
Since $P(2)=P(4)=P(6)$ and $P(E) = P(2)+P(4)+P(6)$, we have $P(2)=P(4)=P(6) = P(E)/3 = (10/19)/3 = 10/57$.
Let $G_3$ be the event that the face value is greater than 3, so $G_3 = \{4, 5, 6\}$. The event $E \cap G_3 = \{4, 6\}$.
$P(E \cap G_3) = P(4) + P(6) = 10/57 + 10/57 = 20/57$.
Given $P(E | G_3) = 0.75$. Using $P(E | G_3) = P(E \cap G_3) / P(G_3)$, we get:
$0.75 = \frac{20/57}{P(G_3)} \implies P(G_3) = \frac{20/57}{0.75} = \frac{20}{57} \cdot \frac{4}{3} = \frac{80}{171}$
$P(G_3) = 80/171 \approx 0.467836$. The closest option is $0.468$.