# 5.2 MCQs-Probability
## Probability
### Basic Concepts
1\. The probability of an event $$E$$, where $$0 \le P(E) \le 1$$, represents:
1. The number of times an event occurs
2. The measure of certainty that the event will occur
3. The ratio of favorable outcomes to total possible outcomes in an experiment
4. Both 2 and 3
Show me the answer
**Answer:** 4. Both 2 and 3
**Explanation:**
* Probability is formally defined as a numerical measure of the likelihood of an event.
* In the classical approach for equally likely outcomes, $$P(E) = \frac{n(E)}{n(S)}$$, where $$n(E)$$ is the number of favorable outcomes and $$n(S)$$ is the total number of possible outcomes in the sample space $$S$$.
* This value always lies between 0 (impossible event) and 1 (certain event).
2\. The sample space for tossing two coins simultaneously is:
1. $${H, T}$$
2. $${HH, HT, TH, TT}$$
3. $${HH, TT}$$
4. $${H, T, H, T}$$
Show me the answer
**Answer:** 2. $${HH, HT, TH, TT}$$
**Explanation:**
* The sample space is the set of all possible distinct outcomes of a random experiment.
* For two coins, each coin can be Head (H) or Tail (T).
* The ordered outcomes are: (H,H), (H,T), (T,H), (T,T).
* Note: $$HT$$ and $$TH$$ are different outcomes if the coins are distinguishable.
* The total number of outcomes is $$2 \times 2 = 4$$.
3\. An event that can never occur has a probability of:
1. 0
2. 1
3. 0.5
4. -1
Show me the answer
**Answer:** 1. 0
**Explanation:**
* The probability of an impossible event is 0.
* For example, in a single die roll, the event "getting a 7" has probability 0.
* Formally, if $$E = \emptyset$$ (empty event), then $$P(E) = 0$$.
* This is an axiom of probability: $$P(\emptyset) = 0$$.
4\. If an event is certain to occur, its probability is:
1. 0
2. 1
3. 0.5
4. Depends on the experiment
Show me the answer
**Answer:** 2. 1
**Explanation:**
* The probability of a certain event is 1.
* For example, in a single die roll, the event "getting a number between 1 and 6 inclusive" has probability 1.
* Formally, if $$E = S$$ (the entire sample space), then $$P(E) = 1$$.
* This is an axiom of probability: $$P(S) = 1$$.
### Calculating Probability
5\. When a fair die is rolled, what is the probability of getting an even number?
1. $$\frac{1}{6}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{2}$$
4. $$\frac{2}{3}$$
Show me the answer
**Answer:** 3. $$\frac{1}{2}$$
**Explanation:**
* Sample space $$S = {1, 2, 3, 4, 5, 6}$$, so $$n(S) = 6$$.
* Event $$E$$: getting an even number = $${2, 4, 6}$$, so $$n(E) = 3$$.
* For a fair die, all outcomes are equally likely.
* $$P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$$.
6\. From a standard deck of 52 cards, one card is drawn at random. What is the probability that it is a King?
1. $$\frac{1}{13}$$
2. $$\frac{1}{4}$$
3. $$\frac{1}{52}$$
4. $$\frac{4}{13}$$
Show me the answer
**Answer:** 1. $$\frac{1}{13}$$
**Explanation:**
* Total cards $$n(S) = 52$$.
* Number of Kings $$n(E) = 4$$.
* $$P(\text{King}) = \frac{4}{52} = \frac{1}{13}$$.
7\. A bag contains 3 red, 4 blue, and 5 green marbles. If one marble is drawn randomly, the probability that it is NOT blue is:
1. $$\frac{4}{12}$$
2. $$\frac{8}{12}$$
3. $$\frac{3}{12}$$
4. $$\frac{5}{12}$$
Show me the answer
**Answer:** 2. $$\frac{8}{12}$$
**Explanation:**
* Total marbles $$n(S) = 3 + 4 + 5 = 12$$.
* Number of blue marbles = 4.
* Event $$E$$: marble is blue.
* We want $$P(\text{Not Blue}) = 1 - P(\text{Blue})$$.
* $$P(\text{Blue}) = \frac{4}{12}$$.
* $$P(\text{Not Blue}) = 1 - \frac{4}{12} = \frac{8}{12} = \frac{2}{3}$$.
* Alternatively, count non-blue marbles: Red (3) + Green (5) = 8. So $$P = \frac{8}{12}$$.
### Complementary Events
8\. If the probability of an event $$A$$ occurring is $$P(A) = 0.7$$, then the probability of its complement $$A'$$ is:
1. 0.3
2. 0.7
3. 0
4. 1
Show me the answer
**Answer:** 1. 0.3
**Explanation:**
* The complement rule states: $$P(A') = 1 - P(A)$$.
* Given $$P(A) = 0.7$$.
* Therefore, $$P(A') = 1 - 0.7 = 0.3$$.
* This follows from the fact that $$A$$ and $$A'$$ are mutually exclusive and exhaustive: $$P(A) + P(A') = 1$$.
9\. For any event $$E$$, which of the following is always true?
1. $$P(E) + P(E') = 0$$
2. $$P(E) + P(E') = 1$$
3. $$P(E) - P(E') = 0$$
4. $$P(E) = P(E')$$
Show me the answer
**Answer:** 2. $$P(E) + P(E') = 1$$
**Explanation:**
* This is the **complement rule** of probability.
* Since $$E$$ and $$E'$$ are mutually exclusive (no common outcomes) and together make up the entire sample space ($$E \cup E' = S$$), their probabilities must sum to 1.
* Formally: $$P(E) + P(E') = P(S) = 1$$.
### Mutually Exclusive Events
10\. Two events $$A$$ and $$B$$ are mutually exclusive if:
1. $$P(A \cap B) = P(A) \times P(B)$$
2. $$P(A \cap B) = 0$$
3. $$P(A \cup B) = P(A) + P(B)$$
4. Both 2 and 3
Show me the answer
**Answer:** 4. Both 2 and 3
**Explanation:**
* Mutually exclusive (disjoint) events cannot occur at the same time.
* This means their intersection is empty: $$A \cap B = \emptyset$$, so $$P(A \cap B) = 0$$.
* For mutually exclusive events, the addition rule simplifies: $$P(A \cup B) = P(A) + P(B)$$.
* If events are not mutually exclusive, the general rule is $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$.
11\. If $$A$$ and $$B$$ are mutually exclusive events with $$P(A) = 0.4$$ and $$P(B) = 0.3$$, then $$P(A \cup B)$$ is:
1. 0.7
2. 0.12
3. 0.1
4. 1.0
Show me the answer
**Answer:** 1. 0.7
**Explanation:**
* For mutually exclusive events, $$P(A \cup B) = P(A) + P(B)$$.
* $$P(A \cup B) = 0.4 + 0.3 = 0.7$$.
* Note: Since they are mutually exclusive, $$P(A \cap B) = 0$$, so it does not need to be subtracted.
### Independent Events
12\. Two events $$A$$ and $$B$$ are independent if:
1. $$P(A \cap B) = 0$$
2. $$P(A \cap B) = P(A) + P(B)$$
3. $$P(A \cap B) = P(A) \times P(B)$$
4. $$P(A \cup B) = P(A) + P(B)$$
Show me the answer
**Answer:** 3. $$P(A \cap B) = P(A) \times P(B)$$
**Explanation:**
* Independence means the occurrence of one event does not affect the probability of the other.
* The mathematical definition is: $$P(A \cap B) = P(A) \cdot P(B)$$.
* This is different from mutually exclusive events, where $$P(A \cap B) = 0$$.
* Two events can be independent but still have some overlap (non-zero intersection).
13\. A coin is tossed twice. The event "first toss is Head" and the event "second toss is Tail" are:
1. Mutually exclusive
2. Independent
3. Both mutually exclusive and independent
4. Neither mutually exclusive nor independent
Show me the answer
**Answer:** 2. Independent
**Explanation:**
* Let A = "first toss is Head", B = "second toss is Tail".
* The outcome of the first toss does not affect the second toss (they are separate trials).
* Therefore, A and B are independent.
* They are NOT mutually exclusive because the outcome (H, T) belongs to both A and B, so $$A \cap B \neq \emptyset$$.
### Conditional Probability
14\. The conditional probability $$P(A|B)$$ is defined as:
1. $$\frac{P(A)}{P(B)}$$
2. $$\frac{P(A \cap B)}{P(B)}$$, provided $$P(B) > 0$$
3. $$\frac{P(B)}{P(A)}$$
4. $$P(A \cap B)$$
Show me the answer
**Answer:** 2. $$\frac{P(A \cap B)}{P(B)}$$, provided $$P(B) > 0$$
**Explanation:**
* Conditional probability is the probability of event A given that event B has occurred.
* The formula is $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$.
* This formula makes sense: we restrict the sample space to event B, and then find the proportion of B's outcomes where A also occurs.
* The condition $$P(B) > 0$$ is necessary to avoid division by zero.
15\. If $$P(A) = 0.5$$, $$P(B) = 0.4$$, and $$P(A \cap B) = 0.2$$, then $$P(A|B)$$ is:
1. 0.1
2. 0.5
3. 0.4
4. 0.8
Show me the answer
**Answer:** 2. 0.5
**Explanation:**
* Using the formula for conditional probability: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = 0.5$$.
* This means, given that B has occurred, the probability of A is 0.5.
16\. For independent events $$A$$ and $$B$$, $$P(A|B)$$ equals:
1. $$P(B)$$
2. $$P(A)$$
3. $$P(A \cap B)$$
4. $$\frac{P(A)}{P(B)}$$
Show me the answer
**Answer:** 2. $$P(A)$$
**Explanation:**
* If A and B are independent, then knowing B occurred gives no information about A.
* Formally, independence means $$P(A \cap B) = P(A)P(B)$$.
* Substituting into the conditional probability formula: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)P(B)}{P(B)} = P(A)$$, provided $$P(B) > 0$$.
### Addition and Multiplication Rules
17\. The general addition rule for any two events $$A$$ and $$B$$ is:
1. $$P(A \cup B) = P(A) + P(B)$$
2. $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
3. $$P(A \cup B) = P(A) \times P(B)$$
4. $$P(A \cup B) = P(A|B) + P(B|A)$$
Show me the answer
**Answer:** 2. $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
**Explanation:**
* When adding probabilities, outcomes in the intersection $$A \cap B$$ are counted twice in $$P(A) + P(B)$$.
* To count them only once, we subtract $$P(A \cap B)$$.
* If A and B are mutually exclusive, then $$P(A \cap B) = 0$$, and the rule simplifies to $$P(A \cup B) = P(A) + P(B)$$.
18\. The multiplication rule for any two events $$A$$ and $$B$$ is:
1. $$P(A \cap B) = P(A) + P(B)$$
2. $$P(A \cap B) = P(A) \times P(B)$$
3. $$P(A \cap B) = P(A) \times P(B|A)$$, provided $$P(A) > 0$$
4. Both 2 and 3
Show me the answer
**Answer:** 3. $$P(A \cap B) = P(A) \times P(B|A)$$, provided $$P(A) > 0$$
**Explanation:**
* The general multiplication rule comes from rearranging the definition of conditional probability: $$P(B|A) = \frac{P(A \cap B)}{P(A)}$$.
* Therefore, $$P(A \cap B) = P(A) \cdot P(B|A)$$.
* Option 2, $$P(A \cap B) = P(A) \times P(B)$$, is only true if A and B are independent.
* The rule can also be written as $$P(A \cap B) = P(B) \cdot P(A|B)$$.
### Law of Total Probability
19\. For events $$A$$ and $$B$$ that form a partition of the sample space ($$A \cup B = S$$ and $$A \cap B = \emptyset$$), and another event $$E$$, the total probability $$P(E)$$ equals:
1. $$P(E|A) + P(E|B)$$
2. $$P(A)P(E|A) + P(B)P(E|B)$$
3. $$P(A) + P(B)$$
4. $$P(E \cap A) + P(E \cap B)$$
Show me the answer
**Answer:** 2. $$P(A)P(E|A) + P(B)P(E|B)$$
**Explanation:**
* The Law of Total Probability states that if $$A\_1, A\_2, ..., A\_n$$ form a partition of $$S$$, then for any event $$E$$: $$P(E) = \sum\_{i=1}^{n} P(A\_i) P(E|A\_i)$$.
* For a partition into two events A and B: $$P(E) = P(A)P(E|A) + P(B)P(E|B)$$.
* This is because $$E = (E \cap A) \cup (E \cap B)$$, and these two parts are disjoint.
### Bayes' Theorem
20\. Bayes' Theorem relates:
1. $$P(A|B)$$ and $$P(B|A)$$
2. $$P(A)$$ and $$P(A')$$
3. $$P(A \cup B)$$ and $$P(A \cap B)$$
4. Marginal and conditional probabilities
Show me the answer
**Answer:** 1. $$P(A|B)$$ and $$P(B|A)$$
**Explanation:**
* Bayes' Theorem provides a way to "reverse" conditional probabilities.
* The formula is: $$P(A|B) = \frac{P(B|A) P(A)}{P(B)}$$.
* It allows us to update the probability of hypothesis $$A$$ given observed evidence $$B$$, using the known probability of evidence $$B$$ given hypothesis $$A$$.
* It is derived from the definition of conditional probability and the multiplication rule.
21\. If $$P(A) = 0.6$$, $$P(B|A) = 0.3$$, and $$P(B|A') = 0.2$$, then using Bayes' Theorem, $$P(A|B)$$ is:
1. $$\frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2}$$
2. $$\frac{0.6 \times 0.3}{0.6 + 0.4}$$
3. $$\frac{0.3}{0.6}$$
4. $$0.6 \times 0.3$$
Show me the answer
**Answer:** 1. $$\frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2}$$
**Explanation:**
* First, find $$P(B)$$ using the Law of Total Probability: $$P(B) = P(A)P(B|A) + P(A')P(B|A')$$.
* Here, $$P(A') = 1 - 0.6 = 0.4$$.
* So, $$P(B) = (0.6 \times 0.3) + (0.4 \times 0.2)$$.
* Now apply Bayes' Theorem: $$P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{0.3 \times 0.6}{(0.6 \times 0.3) + (0.4 \times 0.2)}$$.
### Expected Value
22\. The expected value (mean) of a discrete random variable $$X$$ is defined as:
1. The most frequent value
2. The middle value when sorted
3. $$\sum x\_i P(X = x\_i)$$
4. $$\frac{\sum x\_i}{n}$$
Show me the answer
**Answer:** 3. $$\sum x\_i P(X = x\_i)$$
**Explanation:**
* For a discrete random variable $$X$$ taking values $$x\_i$$ with probabilities $$p\_i$$, the expected value is: $$E\[X] = \sum\_{i} x\_i \cdot p\_i$$.
* It is a weighted average of all possible values, weighted by their probabilities.
* Option 4 describes the sample mean (average of observed values), not the theoretical expected value.
23\. In a game where you win $$$$10 $$with probability 0.1 and lose$$ $$1 $$ with probability 0.9, the expected value of your gain is:
1. $$$$1 $$
2. $$$$0.1 $$
3. $$-$$0.1 $$
4. $$$$0.9 $$
Show me the answer
**Answer:** 2. $$$$0.1 $$
**Explanation:**
* Let $$X$$ be the gain. $$X$$ can be 10 or -1.
* $$E\[X] = (10 \times 0.1) + ((-1) \times 0.9) = 1 - 0.9 = 0.1$$.
* The expected gain is $$$$0.10 $$. This means, on average, you gain 10 cents per play in the long run.
### Variance and Standard Deviation
24\. The variance of a random variable $$X$$ measures:
1. Its average value
2. Its most likely value
3. The spread or dispersion around the mean
4. The probability of extreme values
Show me the answer
**Answer:** 3. The spread or dispersion around the mean
**Explanation:**
* Variance, denoted $$Var(X)$$ or $$\sigma^2$$, quantifies how much the values of $$X$$ differ from the expected value $$E\[X]$$.
* Formula: $$Var(X) = E\[(X - \mu)^2] = \sum (x\_i - \mu)^2 P(X=x\_i)$$, where $$\mu = E\[X]$$.
* A higher variance indicates greater variability.
* The standard deviation is the square root of the variance: $$\sigma = \sqrt{Var(X)}$$.
25\. If $$X$$ is a random variable with $$E\[X] = 5$$ and $$E\[X^2] = 30$$, then its variance is:
1. 5
2. 25
3. 30
4. 35
Show me the answer
**Answer:** 1. 5
**Explanation:**
* A useful formula for variance is: $$Var(X) = E\[X^2] - (E\[X])^2$$.
* Given: $$E\[X] = 5$$, so $$(E\[X])^2 = 25$$.
* Given: $$E\[X^2] = 30$$.
* Therefore, $$Var(X) = 30 - 25 = 5$$.
### Binomial Distribution
26\. Which of the following is NOT a condition for a Binomial experiment?
1. Fixed number of trials (n)
2. Only two possible outcomes per trial (success/failure)
3. Constant probability of success (p) for each trial
4. Trials must be dependent on each other
Show me the answer
**Answer:** 4. Trials must be dependent on each other
**Explanation:**
* The Binomial distribution models the number of successes in $$n$$ independent and identical Bernoulli trials.
* The four conditions are:
1. Fixed number of trials, $$n$$.
2. Each trial has only two outcomes: success (with probability p) or failure (with probability q=1-p).
3. Constant probability of success, p, for each trial.
4. The trials are **independent** of each other (the outcome of one trial does not affect another).
27\. In a Binomial distribution with $$n = 10$$ trials and probability of success $$p = 0.2$$, the mean (expected number of successes) is:
1. 1
2. 2
3. 4
4. 10
Show me the answer
**Answer:** 2. 2
**Explanation:**
* For a Binomial random variable $$X \sim \text{Bin}(n, p)$$:
* Mean: $$\mu = E\[X] = n \cdot p$$
* Variance: $$\sigma^2 = n \cdot p \cdot (1-p)$$
* Here, $$n = 10$$, $$p = 0.2$$.
* $$E\[X] = 10 \times 0.2 = 2$$.
28\. The probability of getting exactly $$k$$ successes in $$n$$ independent trials is given by the formula:
1. $$\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}$$
2. $$n p^k$$
3. $$\frac{p^k}{k!}$$
4. $$\binom{n}{k} p^n (1-p)^k$$
Show me the answer
**Answer:** 1. $$\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}$$
**Explanation:**
* This is the probability mass function (PMF) of the Binomial distribution.
* $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ is the binomial coefficient, counting the number of ways to choose which $$k$$ of the $$n$$ trials are successes.
* $$p^k$$ is the probability of those $$k$$ successes.
* $$(1-p)^{n-k}$$ is the probability of the remaining $$n-k$$ failures.
### Normal Distribution
29\. The standard Normal distribution has:
1. Mean = 0, Variance = 1
2. Mean = 1, Variance = 0
3. Mean = 0, Standard Deviation = 1
4. Both 1 and 3
Show me the answer
**Answer:** 4. Both 1 and 3
**Explanation:**
* The standard Normal distribution, denoted $$Z \sim N(0, 1)$$, is a special case of the Normal distribution.
* It has mean ($$\mu$$) = 0.
* It has variance ($$\sigma^2$$) = 1, which implies standard deviation ($$\sigma$$) = 1.
* Any Normal random variable $$X \sim N(\mu, \sigma^2)$$ can be standardized using $$Z = \frac{X - \mu}{\sigma}$$.
30\. In a Normal distribution, approximately what percentage of data lies within one standard deviation of the mean?
1. 50%
2. 68%
3. 95%
4. 99.7%
Show me the answer
**Answer:** 2. 68%
**Explanation:**
* This is the empirical rule (68-95-99.7 rule) for Normal distributions.
* Approximately:
* 68% of data falls within $$\mu \pm \sigma$$ (one standard deviation).
* 95% of data falls within $$\mu \pm 2\sigma$$.
* 99.7% of data falls within $$\mu \pm 3\sigma$$.
* These are approximate percentages based on the properties of the Normal density curve.