Samacheer Kalvi Class 12 Maths Solutions Chapter 11 Probability Distributions Exercise 11.2

 

Question 1. Three fair coins are tossed simultaneously. Find the probability mass function for a number of heads that occurred.
Answer: When three fair coins are tossed, the possible results are listed in the sample space \( S \). Each outcome has an equal chance of happening. The random variable 'X' represents the number of heads in each outcome.
The sample space is \( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \). The total number of outcomes is 8.
We find the number of outcomes for each possible number of heads:
\( P(X=0) \): 0 heads (TTT) \( \implies 1 \) outcome
\( P(X=1) \): 1 head (HTT, THT, TTH) \( \implies 3 \) outcomes
\( P(X=2) \): 2 heads (HHT, HTH, THH) \( \implies 3 \) outcomes
\( P(X=3) \): 3 heads (HHH) \( \implies 1 \) outcome

The probability for each value of X is calculated by dividing the number of favorable outcomes by the total number of outcomes (8).

The probability mass function \( f(x) \) can be written as:
\[ f(x) = \begin{cases} \frac { 1 }{ 8 }, & \text{for } x = 0, 3 \\ \frac { 3 }{ 8 }, & \text{for } x = 1, 2 \end{cases} \]
In simple words: When you toss three coins, you count how many heads appear. The probability mass function shows how likely each number of heads (0, 1, 2, or 3) is. For 0 or 3 heads, the chance is 1 out of 8. For 1 or 2 heads, the chance is 3 out of 8.

🎯 Exam Tip: Always define your sample space clearly first, then identify the random variable and carefully count the outcomes for each value of the variable to find the probabilities.

 

Question 2. A six-sided die is marked '1' on one face, '3' on two of its faces and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find
(i) the probability mass function
(ii) the cumulative distribution function
(iii) P(4 ≤ X < 10)
(iv) P(X ≥ 6)
Answer: First, let's understand the die. It has 6 faces with scores: one '1', two '3's, and three '5's. The die is thrown two times, and X is the total score. The sample space for the score on a single throw is \( \{1, 3, 3, 5, 5, 5\} \). When the die is thrown twice, the possible total scores are found by adding the scores of the two throws. The table below lists all possible combinations and their sums, which represent the sample space for the total score. The total number of possible outcomes in the sample space \( n(S) = 6 \times 6 = 36 \).

The distinct values of the random variable X (total score) are \( X = \{2, 4, 6, 8, 10\} \). We count the number of times each score appears in the table above:

(i) Probability mass function
The probability mass function \( f(x) = P(X = x) \) for each value of X is found by dividing the number of elements by the total outcomes \( n(S) = 36 \):

(ii) Cumulative distribution function
The cumulative distribution function \( F(x) = P(X \le x) \) sums up probabilities up to a certain value of x.
\( P(X < 2) = 0 \) for \( -\infty < x < 2 \)
\( F(2) = P(X \le 2) = P(X < 2) + P(X = 2) = 0 + \frac { 1 }{ 36 } = \frac { 1 }{ 36 } \)
\( F(4) = P(X \le 4) = P(X \le 2) + P(X = 4) = \frac { 1 }{ 36 } + \frac { 4 }{ 36 } = \frac { 5 }{ 36 } \)
\( F(6) = P(X \le 6) = P(X \le 4) + P(X = 6) = \frac { 5 }{ 36 } + \frac { 10 }{ 36 } = \frac { 15 }{ 36 } \)
\( F(8) = P(X \le 8) = P(X \le 6) + P(X = 8) = \frac { 15 }{ 36 } + \frac { 12 }{ 36 } = \frac { 27 }{ 36 } \)
\( F(10) = P(X \le 10) = P(X \le 8) + P(X = 10) = \frac { 27 }{ 36 } + \frac { 9 }{ 36 } = \frac { 36 }{ 36 } = 1 \)
The cumulative distribution function \( F(x) \) is:
\[ F(x) = \begin{cases} 0, & \text{for } -\infty < x < 2 \\ \frac { 1 }{ 36 }, & \text{for } 2 \le x < 4 \\ \frac { 5 }{ 36 }, & \text{for } 4 \le x < 6 \\ \frac { 15 }{ 36 }, & \text{for } 6 \le x < 8 \\ \frac { 27 }{ 36 }, & \text{for } 8 \le x < 10 \\ 1, & \text{for } 10 \le x < \infty \end{cases} \]
(iii) P(4 ≤ X < 10)
\( P(4 \le X < 10) = P(X = 4) + P(X = 6) + P(X = 8) \)
\( \implies P(4 \le X < 10) = \frac { 4 }{ 36 } + \frac { 10 }{ 36 } + \frac { 12 }{ 36 } = \frac { 26 }{ 36 } = \frac { 13 }{ 18 } \)
(iv) P(X ≥ 6)
\( P(X \ge 6) = P(X = 6) + P(X = 8) + P(X = 10) \)
\( \implies P(X \ge 6) = \frac { 10 }{ 36 } + \frac { 12 }{ 36 } + \frac { 9 }{ 36 } = \frac { 31 }{ 36 } \)
In simple words: First, list all the possible scores you can get from rolling the special die twice. Then, find how many times each total score appears. The probability mass function tells you the chance of getting each exact total score. The cumulative distribution function tells you the chance of getting a score less than or equal to a certain value. We then use these probabilities to find specific range probabilities.

🎯 Exam Tip: When dealing with custom dice or multiple trials, always construct a clear table or list of all possible outcomes to accurately count frequencies for each sum or event.

 

Question 3. Find the probability mass function and cumulative distribution function of a number of girl children in families with 4 children, assuming equal probabilities for boys and girls.
Answer: Let X be the random variable representing the number of girl children among 4 children. Since boys and girls have equal probabilities, \( P(\text{Boy}) = P(\text{Girl}) = \frac { 1 }{ 2 } \). The possible values for X are \( \{0, 1, 2, 3, 4\} \). This scenario follows a binomial distribution. We can find the number of ways to have 'x' girls out of 4 children using combinations \( \binom{4}{x} \). For 4 children, the total number of possible gender arrangements is \( 2^4 = 16 \). - 0 girls (BBBB): \( \binom{4}{0} = 1 \) outcome - 1 girl (GBBB, BGBB, BBGB, BBBG): \( \binom{4}{1} = 4 \) outcomes - 2 girls (GGBB, GBGB, GBBG, BGGB, BGBG, BBGG): \( \binom{4}{2} = 6 \) outcomes - 3 girls (GGGB, GGBG, GBGG, BGGG): \( \binom{4}{3} = 4 \) outcomes - 4 girls (GGGG): \( \binom{4}{4} = 1 \) outcome

(i) Probability mass function
The probability mass function \( f(x) = P(X = x) \) is found by dividing the number of elements by the total possible arrangements (16):

(ii) Cumulative distribution function
The cumulative distribution function \( F(x) = P(X \le x) \) sums up the probabilities.
\( P(X < 0) = 0 \) for \( -\infty < x < 0 \)
\( F(0) = P(X \le 0) = P(X = 0) = \frac { 1 }{ 16 } \)
\( F(1) = P(X \le 1) = P(X = 0) + P(X = 1) = \frac { 1 }{ 16 } + \frac { 4 }{ 16 } = \frac { 5 }{ 16 } \)
\( F(2) = P(X \le 2) = P(X \le 1) + P(X = 2) = \frac { 5 }{ 16 } + \frac { 6 }{ 16 } = \frac { 11 }{ 16 } \)
\( F(3) = P(X \le 3) = P(X \le 2) + P(X = 3) = \frac { 11 }{ 16 } + \frac { 4 }{ 16 } = \frac { 15 }{ 16 } \)
\( F(4) = P(X \le 4) = P(X \le 3) + P(X = 4) = \frac { 15 }{ 16 } + \frac { 1 }{ 16 } = \frac { 16 }{ 16 } = 1 \)
The cumulative distribution function \( F(x) \) is:
\[ F(x) = \begin{cases} 0, & \text{for } x < 0 \\ \frac { 1 }{ 16 }, & \text{for } 0 \le x < 1 \\ \frac { 5 }{ 16 }, & \text{for } 1 \le x < 2 \\ \frac { 11 }{ 16 }, & \text{for } 2 \le x < 3 \\ \frac { 15 }{ 16 }, & \text{for } 3 \le x < 4 \\ 1, & \text{for } 4 \le x \end{cases} \]
In simple words: We are looking at families with four children and assuming each child is equally likely to be a boy or a girl. The probability mass function tells us the exact chance of having 0, 1, 2, 3, or 4 girls. The cumulative distribution function then shows the total chance of having up to a certain number of girls. This is a common setup for binomial probability, where there are only two outcomes for each trial.

🎯 Exam Tip: When dealing with "equal probabilities" in scenarios like coin tosses or gender, remember to use combinations to count outcomes for the probability mass function. Sum all probabilities to 1 to check your work.

 

Question 4. Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by
\[ f(x) = \begin{cases} \frac { x^2+1 }{ k }, & \text{for } x = 0,1,2 \\ 0, & \text{otherwise} \end{cases} \]
Find (i) the value of k.
(ii) cumulative distribution function
(iii) P(X ≥ 1)
Answer: For a probability mass function (PMF), the sum of all probabilities for all possible values of X must be equal to 1. Here, X can take values 0, 1, and 2.

(i) Value of k
We calculate \( f(x) \) for each value of x:
\( f(0) = \frac { 0^2+1 }{ k } = \frac { 1 }{ k } \)
\( f(1) = \frac { 1^2+1 }{ k } = \frac { 2 }{ k } \)
\( f(2) = \frac { 2^2+1 }{ k } = \frac { 5 }{ k } \)
The probability mass function table is:

Sum of probabilities must be 1:
\( \sum_{x} f(x) = 1 \)
\( \frac { 1 }{ k } + \frac { 2 }{ k } + \frac { 5 }{ k } = 1 \)
\( \frac { 1+2+5 }{ k } = 1 \)
\( \frac { 8 }{ k } = 1 \)
\( \implies k = 8 \)

(ii) Cumulative distribution function
Now that we know \( k = 8 \), the PMF values are:
\( f(0) = \frac { 1 }{ 8 } \)
\( f(1) = \frac { 2 }{ 8 } \)
\( f(2) = \frac { 5 }{ 8 } \)
The cumulative distribution function \( F(x) = P(X \le x) \) is calculated as follows:
\( F(0) = P(X \le 0) = P(X = 0) = \frac { 1 }{ 8 } \)
\( F(1) = P(X \le 1) = P(X = 0) + P(X = 1) = \frac { 1 }{ 8 } + \frac { 2 }{ 8 } = \frac { 3 }{ 8 } \)
\( F(2) = P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac { 1 }{ 8 } + \frac { 2 }{ 8 } + \frac { 5 }{ 8 } = 1 \)
The cumulative distribution function table is:

(iii) P(X ≥ 1)
\( P(X \ge 1) = P(X = 1) + P(X = 2) \)
\( \implies P(X \ge 1) = \frac { 2 }{ 8 } + \frac { 5 }{ 8 } = \frac { 7 }{ 8 } \)
In simple words: For any list of probabilities to be valid, they must all be positive and add up to 1. We used this rule to find the missing value 'k'. Once 'k' is known, we can find the exact chance of each outcome and then figure out the chance of getting a value greater than or equal to 1.

🎯 Exam Tip: Always remember that the sum of all probabilities in a probability mass function must equal 1. This is the key property used to find unknown constants like 'k'.

 

Question 5. The cumulative distribution function of a discrete random variable is given by
\[ F(x) = \begin{cases} 0 & -\infty < x < -1 \\ 0.15 & -1 \le x < 0 \\ 0.35 & 0 \le x < 1 \\ 0.60 & 1 \le x < 2 \\ 0.85 & 2 \le x < 3 \\ 1 & 3 \le x < \infty \end{cases} \]
Find (i) the probability mass function
(ii) P(X < 1)
(iii) P(X ≥ 2)
Answer: The cumulative distribution function (CDF), \( F(x) \), tells us the probability that the random variable X takes a value less than or equal to x. For a discrete random variable, the probability mass function (PMF), \( f(x) \), can be found by subtracting consecutive CDF values: \( f(x) = F(x) - F(x-1) \). The points where \( F(x) \) jumps are the values X can take.

(i) The probability mass function
The jumps in \( F(x) \) occur at \( x = -1, 0, 1, 2, 3 \). These are the possible values for X.
\( f(-1) = P(X = -1) = F(-1) - F(\text{value just before -1}) = F(-1) - F(\text{any } x < -1) = 0.15 - 0 = 0.15 \)
\( f(0) = P(X = 0) = F(0) - F(\text{value just before 0}) = F(0) - F(-1) = 0.35 - 0.15 = 0.20 \)
\( f(1) = P(X = 1) = F(1) - F(\text{value just before 1}) = F(1) - F(0) = 0.60 - 0.35 = 0.25 \)
\( f(2) = P(X = 2) = F(2) - F(\text{value just before 2}) = F(2) - F(1) = 0.85 - 0.60 = 0.25 \)
\( f(3) = P(X = 3) = F(3) - F(\text{value just before 3}) = F(3) - F(2) = 1 - 0.85 = 0.15 \)

(ii) P(X < 1)
\( P(X < 1) \) means the probability that X is less than 1. This includes values \( X = -1 \) and \( X = 0 \).
\( P(X < 1) = P(X = -1) + P(X = 0) \)
\( \implies P(X < 1) = 0.15 + 0.20 = 0.35 \)

(iii) P(X ≥ 2)
\( P(X \ge 2) \) means the probability that X is greater than or equal to 2. This includes values \( X = 2 \) and \( X = 3 \).
\( P(X \ge 2) = P(X = 2) + P(X = 3) \)
\( \implies P(X \ge 2) = 0.25 + 0.15 = 0.40 \)
In simple words: The cumulative distribution function builds up probabilities, showing the chance of being at or below a certain point. To find the probability of a single exact point, we look at how much the cumulative probability increases at that point. Once we have the probabilities for each specific value, we can add them up for any range we need.

🎯 Exam Tip: Remember that for a discrete CDF, \( f(x) = F(x) - F(x_{\text{previous}}) \). For \( P(a \le X < b) \), sum \( f(x) \) for all x values in that range, or use \( F(b-1) - F(a-1) \) if \( a \) and \( b \) are integers.

 

Question 6. A random variable X has the following probability mass function.

Find (i) the value of k
(ii) P(2 ≤ X < 5)
(iii) P(3 < X)
Answer: For a function to be a valid probability mass function, two conditions must be met: all probabilities \( f(x) \) must be non-negative, and the sum of all probabilities must be equal to 1. We will use the second condition to find the value of k. The provided solution assumes the probabilities are \( k^2, 2k^2, 3k^2, 2k, 3k \) in its calculation of k.

(i) The value of k
Sum of probabilities \( = 1 \):
\( k^2 + 2k^2 + 3k^2 + 2k + 3k = 1 \)
\( 6k^2 + 5k = 1 \)
\( 6k^2 + 5k - 1 = 0 \)
We can solve this quadratic equation by factoring:
\( 6k^2 + 6k - k - 1 = 0 \)
\( 6k(k + 1) - 1(k + 1) = 0 \)
\( (6k - 1)(k + 1) = 0 \)
This gives two possible values for k: \( k = \frac { 1 }{ 6 } \) or \( k = -1 \).
Since probabilities cannot be negative, we must have \( k = \frac { 1 }{ 6 } \) (we neglect the negative term \( k = -1 \)).

Now, with \( k = \frac { 1 }{ 6 } \), the probability mass function values are (as presented in the source after finding k):

To verify the above table, if we assume \( f(1)=k \), \( f(2)=2k \), \( f(3)=3k \), \( f(4)=2k \) and \( f(5)=3k \) and \( k=1/6 \), then we should have \( f(1) = 1/6, f(2)=2/6, f(3)=3/6, f(4)=2/6, f(5)=3/6 \). However, the table uses \( 1/36, 2/36, 3/36 \) for \( x=1,2,3 \) and \( 2/6, 3/6 \) for \( x=4,5 \). The values in the solution's table (after finding k) are derived from the idea that \( P(x) = \frac{x}{6 \times 6} = \frac{x}{36} \) for \(x=1,2,3\) and \( P(x) = \frac{x}{6} \) for \(x=4,5\). Let's use the probabilities from the presented table for the remaining calculations:
\( f(1) = \frac { 1 }{ 36 } \), \( f(2) = \frac { 2 }{ 36 } \), \( f(3) = \frac { 3 }{ 36 } \), \( f(4) = \frac { 12 }{ 36 } \), \( f(5) = \frac { 18 }{ 36 } \).

(ii) P(2 ≤ X < 5)
This means we need to find the probability that X is 2, 3, or 4.
\( P(2 \le X < 5) = P(X = 2) + P(X = 3) + P(X = 4) \)
\( \implies P(2 \le X < 5) = \frac { 2 }{ 36 } + \frac { 3 }{ 36 } + \frac { 12 }{ 36 } \)
\( \implies P(2 \le X < 5) = \frac { 2+3+12 }{ 36 } = \frac { 17 }{ 36 } \)

(iii) P(3 < X)
This means we need to find the probability that X is greater than 3. This includes values \( X = 4 \) and \( X = 5 \).
\( P(3 < X) = P(X = 4) + P(X = 5) \)
\( \implies P(3 < X) = \frac { 12 }{ 36 } + \frac { 18 }{ 36 } \)
\( \implies P(3 < X) = \frac { 12+18 }{ 36 } = \frac { 30 }{ 36 } = \frac { 5 }{ 6 } \)
In simple words: For any set of probabilities to be real, they must add up to one. We used this rule to find the hidden value of 'k'. Once 'k' is known, we can figure out the chance of X being within certain ranges by simply adding up the probabilities for those specific values of X.

🎯 Exam Tip: When faced with discrepancies between problem statement and solution calculations, always follow the internal consistency of the *provided solution's steps*. Clearly state any derived PMF values used for subsequent calculations.

 

Question 7. The cumulative distribution function of a discrete random variable is given by
\[ F(x) = \begin{cases} 0 & -\infty < x < 0 \\ \frac { 1 }{ 5 } & 0 \le x < 1 \\ \frac { 2 }{ 5 } & 1 \le x < 2 \\ \frac { 3 }{ 5 } & 2 \le x < 3 \\ \frac { 9 }{ 10 } & 3 \le x < 4 \\ 1 & 4 \le x < \infty \end{cases} \]
Find (i) the probability mass function
(ii) P(X < 3) and (iii) P(X ≥ 2)
Answer: The cumulative distribution function (CDF), \( F(x) \), tells us the probability that the random variable X takes a value less than or equal to x. For a discrete random variable, the probability mass function (PMF), \( f(x) \), can be found by looking at the jumps in the CDF: \( f(x) = F(x) - F(x_{\text{previous}}) \). The values of x where \( F(x) \) changes are the possible values that X can take.

(i) The probability mass function
The jumps in \( F(x) \) occur at \( x = 0, 1, 2, 3, 4 \). These are the possible values for X.
\( f(0) = P(X = 0) = F(0) - F(\text{value just before 0}) = \frac { 1 }{ 5 } - 0 = \frac { 1 }{ 5 } \)
\( f(1) = P(X = 1) = F(1) - F(0) = \frac { 2 }{ 5 } - \frac { 1 }{ 5 } = \frac { 1 }{ 5 } \)
\( f(2) = P(X = 2) = F(2) - F(1) = \frac { 3 }{ 5 } - \frac { 2 }{ 5 } = \frac { 1 }{ 5 } \)
\( f(3) = P(X = 3) = F(3) - F(2) = \frac { 9 }{ 10 } - \frac { 3 }{ 5 } = \frac { 9 }{ 10 } - \frac { 6 }{ 10 } = \frac { 3 }{ 10 } \)
\( f(4) = P(X = 4) = F(4) - F(3) = 1 - \frac { 9 }{ 10 } = \frac { 1 }{ 10 } \)
The probability mass function table is:

(ii) P(X < 3)
\( P(X < 3) \) means the probability that X is less than 3. This includes values \( X = 0, 1, 2 \).
\( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \)
\( \implies P(X < 3) = \frac { 1 }{ 5 } + \frac { 1 }{ 5 } + \frac { 1 }{ 5 } = \frac { 3 }{ 5 } \)
Alternatively, using the CDF: \( P(X < 3) = F(\text{value just before 3}) = F(2.99...) = 0.60 = \frac{3}{5} \).

(iii) P(X ≥ 2)
\( P(X \ge 2) \) means the probability that X is greater than or equal to 2. This includes values \( X = 2, 3, 4 \).
\( P(X \ge 2) = P(X = 2) + P(X = 3) + P(X = 4) \)
\( \implies P(X \ge 2) = \frac { 1 }{ 5 } + \frac { 3 }{ 10 } + \frac { 1 }{ 10 } \)
\( \implies P(X \ge 2) = \frac { 2 }{ 10 } + \frac { 3 }{ 10 } + \frac { 1 }{ 10 } = \frac { 6 }{ 10 } = \frac { 3 }{ 5 } \)
Alternatively, using the CDF: \( P(X \ge 2) = 1 - P(X < 2) = 1 - F(\text{value just before 2}) = 1 - F(1.99...) = 1 - F(1) = 1 - \frac{2}{5} = \frac{3}{5} \).
In simple words: We start with a cumulative distribution function (CDF) which shows the probability of a value being less than or equal to a certain number. We then use this to find the probability mass function (PMF), which gives the exact chance of each individual number occurring. After that, we can easily calculate probabilities for different ranges of X by adding the individual probabilities or by using the properties of the CDF.

🎯 Exam Tip: Remember that for a discrete random variable, \( P(X \ge a) = 1 - P(X < a) = 1 - F(a_{\text{previous}}) \) and \( P(X < a) = F(a_{\text{previous}}) \). These rules simplify calculations when working with CDFs.

 

Question 7. The cumulative distribution function of a discrete random variable is given by
\[ F(x) = \begin{cases} 0 & \text{for } -\infty < x < 0 \\ \frac{1}{2} & \text{for } 0 \le x < 1 \\ \frac{3}{5} & \text{for } 1 \le x < 2 \\ \frac{4}{5} & \text{for } 2 \le x < 3 \\ \frac{9}{10} & \text{for } 3 \le x < 4 \\ 1 & \text{for } 4 \le x < \infty \end{cases} \]
Find
(i) the probability mass function
(ii) P(X < 3) and (iii) P(X ≥ 2)
Answer:
(i) To find the probability mass function \( f(x) \), we use the property \( f(x_i) = F(x_i) - F(x_{i-1}) \). This means we subtract the cumulative probability of the previous value from the current one.

\( f(0) = F(0) - F(x < 0) = \frac{1}{2} - 0 = \frac{1}{2} \)
\( f(1) = F(1) - F(0) = \frac{3}{5} - \frac{1}{2} = \frac{6}{10} - \frac{5}{10} = \frac{1}{10} \)
\( f(2) = F(2) - F(1) = \frac{4}{5} - \frac{3}{5} = \frac{1}{5} \)
\( f(3) = F(3) - F(2) = \frac{9}{10} - \frac{4}{5} = \frac{9}{10} - \frac{8}{10} = \frac{1}{10} \)
\( f(4) = F(4) - F(3) = 1 - \frac{9}{10} = \frac{1}{10} \)

The probability mass function is:

(ii) To find \( P(X < 3) \), we sum the probabilities for \( X=0, 1, \) and \( 2 \). This is simply \( F(2) \) since \( F(x) \) gives the probability that \( X \) is less than or equal to \( x \). So, \( P(X < 3) \) includes all values up to \( X=2 \).

\( P(X < 3) = P(X=0) + P(X=1) + P(X=2) \)
\( = \frac{1}{2} + \frac{1}{10} + \frac{1}{5} \)
\( = \frac{5}{10} + \frac{1}{10} + \frac{2}{10} = \frac{8}{10} = \frac{4}{5} \)

(iii) To find \( P(X \ge 2) \), we sum the probabilities for \( X=2, 3, \) and \( 4 \). This represents the chance of getting a value of 2 or higher. We can also find this by subtracting \( P(X < 2) \) from 1.

\( P(X \ge 2) = P(X=2) + P(X=3) + P(X=4) \)
\( = \frac{1}{5} + \frac{1}{10} + \frac{1}{10} \)
\( = \frac{2}{10} + \frac{1}{10} + \frac{1}{10} = \frac{4}{10} = \frac{2}{5} \)
In simple words: First, we find the individual probabilities for each outcome using the given cumulative distribution function. Then, to find the probability of X being less than 3, we add up the probabilities for X=0, 1, and 2. To find the probability of X being 2 or more, we add the probabilities for X=2, 3, and 4. Remember that a cumulative distribution function always increases or stays the same as X increases.

🎯 Exam Tip: Always remember that for a discrete random variable, \( P(X \le x) = F(x) \) and \( P(X < x) = F(x-1) \). Also, \( P(a \le X \le b) = F(b) - F(a-1) \). Carefully check the strict inequality signs.