Binomial Distribution - IB Mathematics Analysis and Approaches

1. For a random variable $X \sim B(4, 1/3)$, find the value of $P(X=2 | X \ge 1)$.
   Answer: $\frac{24}{65}$ The conditional probability is given by $P(X=2 | X \ge 1) = \frac{P(X=2 \text{ and } X \ge 1)}{P(X \ge 1)}$. Since the event $X=2$ is a subset of the event $X \ge 1$, their intersection is just $X=2$. So we need to calculate $\frac{P(X=2)}{P(X \ge 1)}$. First, $P(X \ge 1) = 1 - P(X=0)$. $P(X=0) = \binom{4}{0} (1/3)^0 (2/3)^4 = (2/3)^4 = 16/81$. So, $P(X \ge 1) = 1 - 16/81 = 65/81$. Next, $P(X=2) = \binom{4}{2} (1/3)^2 (2/3)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81}$. Therefore, $P(X=2 | X \ge 1) = \frac{24/81}{65/81} = \frac{24}{65}$.
2. For a random variable $X \sim B(3, 1/4)$, what is the probability that the number of successes is an odd number?
   Answer: $\frac{28}{64}$ We need to find the probability that $X$ is odd, which for $n=3$ means $X=1$ or $X=3$. We calculate $P(X=1) + P(X=3)$. $P(X=1) = \binom{3}{1} (1/4)^1 (3/4)^2 = 3 \cdot \frac{1}{4} \cdot \frac{9}{16} = \frac{27}{64}$. $P(X=3) = \binom{3}{3} (1/4)^3 (3/4)^0 = 1 \cdot \frac{1}{64} = \frac{1}{64}$. The total probability is $\frac{27}{64} + \frac{1}{64} = \frac{28}{64}$, which simplifies to $7/16$.
3. Let $X \sim B(4, p)$. Given that the mean of $X$ is 1, find the probability that the number of successes is an extreme outcome (either 0 or 4).
   Answer: $\frac{82}{256}$ First, find the value of $p$. The mean is $E(X) = np = 1$. Given $n=4$, we have $4p = 1$, so $p = 1/4$. We need to find $P(X=0) + P(X=4)$. $P(X=0) = \binom{4}{0} (1/4)^0 (3/4)^4 = (3/4)^4 = \frac{81}{256}$. $P(X=4) = \binom{4}{4} (1/4)^4 (3/4)^0 = (1/4)^4 = \frac{1}{256}$. The sum is $P(X=0) + P(X=4) = \frac{81}{256} + \frac{1}{256} = \frac{82}{256}$.
4. For a random variable $X \sim B(15, 2/5)$, for which integer value of $k$ is the probability $P(X=k)$ at its maximum?
   Answer: 6 The value of $k$ that maximizes $P(X=k)$ is the mode of the distribution. For a binomial distribution $B(n, p)$, the mode is given by $\lfloor (n+1)p \rfloor$. Here, $n=15$ and $p=2/5=0.4$. We calculate $(n+1)p = (15+1)(2/5) = 16 \cdot \frac{2}{5} = \frac{32}{5} = 6.4$. Since 6.4 is not an integer, the mode is unique and is equal to $\lfloor 6.4 \rfloor = 6$. The mean is $np = 15(2/5) = 6$. The mode is always the integer(s) closest to the mean, which in this case is 6.
5. In a city, the probability of rain on any given day is $1/3$, independently of other days. What is the probability that in a 4-day period, there is at most one day of rain?
   Answer: $\frac{48}{81}$ Let $X$ be the number of rainy days in the 4-day period. Then $X \sim B(4, 1/3)$. 'At most one day of rain' means $X=0$ or $X=1$. We need to find $P(X \le 1) = P(X=0) + P(X=1)$. First, $P(X=0) = \binom{4}{0}(1/3)^0(2/3)^4 = (2/3)^4 = \frac{16}{81}$. Next, $P(X=1) = \binom{4}{1}(1/3)^1(2/3)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81}$. The total probability is $P(X \le 1) = \frac{16}{81} + \frac{32}{81} = \frac{48}{81}$.
6. Let the random variable $X$ follow a binomial distribution, $X \sim B(n, p)$. The mean of $X$ is 6 and the variance is 4. Find the value of $P(X=5)$.
   Answer: $\binom{18}{5} (\frac{1}{3})^5 (\frac{2}{3})^{13}$ The mean of a binomial distribution is given by $E(X) = np$ and the variance by $Var(X) = np(1-p)$. Given $np=6$ and $np(1-p)=4$, we can substitute the first equation into the second: $6(1-p) = 4 \Rightarrow 1-p = 4/6 = 2/3 \Rightarrow p = 1/3$. Now, we find $n$: $n(1/3) = 6 \Rightarrow n = 18$. So, $X \sim B(18, 1/3)$. We need to find $P(X=5)$: $P(X=5) = \binom{18}{5} (1/3)^5 (2/3)^{18-5} = \binom{18}{5} (\frac{1}{3})^5 (\frac{2}{3})^{13}$.
7. An archer hits a target with probability $p=2/5$. All shots are independent. If the archer takes 4 shots, what is the probability that they hit the target on exactly two of the first three shots, and also hit the target on the fourth shot?
   Answer: $\frac{72}{625}$ This problem has two independent parts. First, the probability of hitting exactly two of the first three shots. Let $Y$ be the number of hits in the first three shots, so $Y \sim B(3, 2/5)$. We need $P(Y=2)$. $P(Y=2) = \binom{3}{2} (2/5)^2 (3/5)^1 = 3 \cdot \frac{4}{25} \cdot \frac{3}{5} = \frac{36}{125}$. Second, the probability of hitting the target on the fourth shot is simply $p=2/5$. Since the shots are independent, we multiply the probabilities of these two events: Total Probability = $P(Y=2) \times P(\text{hit on 4th}) = \frac{36}{125} \times \frac{2}{5} = \frac{72}{625}$.
8. Consider two binomial random variables, $X \sim B(10, 0.1)$ and $Y \sim B(10, 0.6)$. Which of the following statements is correct?
   Answer: The mode of $X$ is less than the mode of $Y$. Let's evaluate each statement. A: Skewness of a binomial distribution is determined by how far $p$ is from 0.5. For $X$, $|0.1 - 0.5| = 0.4$. For $Y$, $|0.6 - 0.5| = 0.1$. Since $0.4 > 0.1$, the distribution of $X$ is more skewed than $Y$. So A is false. B: $Var(X) = 10(0.1)(0.9) = 0.9$. $Var(Y) = 10(0.6)(0.4) = 2.4$. Since $0.9 < 2.4$, B is false. C: $E(X) = 10(0.1) = 1$. $E(Y) = 10(0.6) = 6$. Since $1 < 6$, C is false. D: Mode of $X$ is $\lfloor (10+1)(0.1) \rfloor = \lfloor 1.1 \rfloor = 1$. Mode of $Y$ is $\lfloor (10+1)(0.6) \rfloor = \lfloor 6.6 \rfloor = 6$. Since $1 < 6$, D is correct.
9. A fair six-sided die is rolled 4 times. Let $X$ be the number of times a '6' is rolled. Given that at least one '6' is rolled, what is the probability that exactly two '6's are rolled?
   Answer: $\frac{150}{671}$ This is a conditional probability problem involving a binomial distribution. When you see "given that" in a probability question, you need to use the conditional probability formula: P(A|B) = P(A and B) / P(B). Let's define our events clearly. You want P(exactly 2 sixes | at least 1 six). Since X follows a binomial distribution with n=4 trials and p=1/6 probability of success, you can calculate the required probabilities. First, find P(exactly 2 sixes): $P(X=2) = \binom{4}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 = 6 \cdot \frac{1}{36} \cdot \frac{25}{36} = \frac{150}{1296}$ Next, find P(at least 1 six). It's easier to use the complement: $P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296}$ Since "exactly 2 sixes" is already contained within "at least 1 six," P(exactly 2 sixes AND at least 1 six) = P(exactly 2 sixes). Therefore: $P(X=2|X \geq 1) = \frac{150/1296}{671/1296} = \frac{150}{671}$ This confirms answer A is correct. Answer B gives the unconditional probability but with wrong calculations. Answer C is just P(X=2) without applying the conditional probability formula. Answer D uses an incorrect denominator, possibly confusing this with a different probability calculation. Remember: conditional probability problems require you to restrict your sample space to the given condition, which always changes the denominator of your final calculation.
10. Let $X \sim B(4, 1/3)$. Find the probability $P(1 \le X < 3)$.
    Answer: $\frac{56}{81}$ We need to find $P(1 \le X < 3)$, which is equal to $P(X=1) + P(X=2)$. First, calculate $P(X=1)$: $P(X=1) = \binom{4}{1} (1/3)^1 (2/3)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81}$. Next, calculate $P(X=2)$: $P(X=2) = \binom{4}{2} (1/3)^2 (2/3)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81}$. Now, add the probabilities: $P(1 \le X < 3) = \frac{32}{81} + \frac{24}{81} = \frac{56}{81}$.
11. Let $X \sim B(13, 0.4)$. Find the mode of the distribution of $X$, which is the most likely number of successes.
    Answer: 5 The mode of a binomial distribution $B(n, p)$ is the integer value $k$ that maximizes $P(X=k)$. The mode is given by $\lfloor (n+1)p \rfloor$. If $(n+1)p$ is an integer, then there are two modes: $(n+1)p$ and $(n+1)p - 1$. In this case, $n=13$ and $p=0.4$. $(n+1)p = (13+1)(0.4) = 14(0.4) = 5.6$. Since 5.6 is not an integer, the mode is unique and is given by $\lfloor 5.6 \rfloor = 5$. The mean of the distribution is $np = 13(0.4) = 5.2$, and the mode is always close to the mean.
12. Let $X \sim B(n, p)$. Given that $P(X=0) = \frac{1}{256}$ and $P(X=1) = \frac{8}{256}$, find the value of $n$.
    Answer: 8 We are given two equations: 1) $P(X=0) = (1-p)^n = \frac{1}{256}$ 2) $P(X=1) = np(1-p)^{n-1} = \frac{8}{256}$ Divide equation (2) by equation (1): $\frac{np(1-p)^{n-1}}{(1-p)^n} = \frac{8/256}{1/256} = 8$ which simplifies to $\frac{np}{1-p} = 8$. From this, we have $np = 8(1-p) \Rightarrow p = \frac{8}{n+8}$ and $1-p = \frac{n}{n+8}$. Substitute $1-p$ back into equation (1): $(\frac{n}{n+8})^n = \frac{1}{256}$. We can test the integer options for $n$. Let's try $n=8$: $(\frac{8}{8+8})^8 = (\frac{8}{16})^8 = (\frac{1}{2})^8 = \frac{1}{256}$. This matches the given probability. Thus, $n=8$.
13. The number of successful outcomes in a sequence of trials is modelled by $X \sim B(25, p)$. The variance of $X$ is 6. Given that the probability of success is less than the probability of failure, find the mean of $X$.
    Answer: 10 The variance is $Var(X) = np(1-p) = 6$. With $n=25$, we have $25p(1-p) = 6$. This leads to the quadratic equation $25p^2 - 25p + 6 = 0$. Factoring gives $(5p-2)(5p-3) = 0$, so the possible values for $p$ are $p=2/5$ and $p=3/5$. The condition 'probability of success is less than the probability of failure' means $p < 1-p$, which simplifies to $2p < 1$ or $p < 0.5$. We must choose $p=2/5 = 0.4$. The mean is $E(X) = np = 25 \cdot (2/5) = 10$.
14. Let $X \sim B(n, p)$. The mean of the distribution is 0.5. Express $P(X \ge 1)$ in terms of $n$.
    Answer: $1 - (\frac{2n-1}{2n})^n$ When you encounter a binomial distribution problem where you need to find probabilities using given parameters, start by extracting all the information from the mean to determine the relationship between variables. Given that $X \sim B(n, p)$ with mean = 0.5, you can use the fact that for a binomial distribution, the mean equals $np$. Therefore: $np = 0.5$, which gives us $p = \frac{0.5}{n} = \frac{1}{2n}$. To find $P(X \geq 1)$, use the complement rule: $P(X \geq 1) = 1 - P(X = 0)$. For a binomial distribution, $P(X = 0) = \binom{n}{0}p^0(1-p)^n = (1-p)^n$. Substituting $p = \frac{1}{2n}$: $P(X = 0) = \left(1 - \frac{1}{2n}\right)^n = \left(\frac{2n-1}{2n}\right)^n$ Therefore: $P(X \geq 1) = 1 - \left(\frac{2n-1}{2n}\right)^n$ This confirms answer A is correct. Looking at the wrong answers: B gives $1 - \left(\frac{1}{2n}\right)^n$, which would result from incorrectly using $p^n$ instead of $(1-p)^n$ for $P(X = 0)$. C shows $1 - \left(\frac{n-1}{n}\right)^n$, suggesting confusion about the relationship between $n$ and $p$. D gives $1 - \left(\frac{1}{2}\right)^n$, which ignores the derived value of $p$ entirely and assumes $p = \frac{1}{2}$. Study tip: Always use the complement rule for "at least" problems in binomial distributions—it's much simpler than calculating multiple individual probabilities. Remember that the mean constraint gives you crucial information about the parameters' relationship.
15. A fair coin is tossed 10 times. Let $X$ be the number of heads obtained. Which of the following statements is false?
    Answer: The variance of the number of heads is 5. For a fair coin tossed 10 times, we have a binomial distribution $X \sim B(10, 0.5)$. A: The expected value is $E(X) = np = 10(0.5) = 5$. This statement is true. B: For $p=0.5$, the distribution is symmetric about the mean. Thus, $P(X=k) = P(X=n-k)$. For $k=4$, $P(X=4) = P(X=10-4) = P(X=6)$. This statement is true. C: The variance is $Var(X) = np(1-p) = 10(0.5)(0.5) = 2.5$. The statement claims the variance is 5, which is false. D: By symmetry, the probability of being $k$ units or more below the mean is the same as being $k$ units or more above the mean. $P(X \le 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)$. $P(X \ge 7) = P(X=7)+P(X=8)+P(X=9)+P(X=10)$. By symmetry $P(X=3)=P(X=7)$, $P(X=2)=P(X=8)$, etc. So the sums are equal. This statement is true.
16. For a random variable $X \sim B(2, p)$, it is given that $P(X=1) = 4/9$. Given that $p > 0.5$, find the expected value of $X$.
    Answer: $4/3$ The probability mass function gives $P(X=1) = \binom{2}{1} p^1 (1-p)^1 = 2p(1-p)$. We are given $P(X=1) = 4/9$, so: $2p(1-p) = 4/9 \Rightarrow p(1-p) = 2/9$. $p - p^2 = 2/9 \Rightarrow 9p - 9p^2 = 2 \Rightarrow 9p^2 - 9p + 2 = 0$. Factoring: $(3p-1)(3p-2) = 0$, so $p = 1/3$ or $p = 2/3$. Since we are given that $p > 0.5$, we choose $p = 2/3$. The expected value is $E(X) = np = 2 \cdot (2/3) = 4/3$.
17. For a random variable $X \sim B(n, 1/4)$, it is found that $P(X=2) = \frac{9}{2} \times P(X=1)$. Find the value of $n$.
    Answer: 28 We are given the relation $P(X=2) = \frac{9}{2} P(X=1)$. Let's write this out using the binomial formula: $\binom{n}{2} (1/4)^2 (3/4)^{n-2} = \frac{9}{2} \cdot \binom{n}{1} (1/4)^1 (3/4)^{n-1}$. $\frac{n(n-1)}{2} \cdot \frac{1}{16} \cdot (3/4)^{n-2} = \frac{9}{2} \cdot n \cdot \frac{1}{4} \cdot (3/4)^{n-1}$. Assuming $n \ge 2$, we can divide by $n$ and other common terms. To simplify, divide both sides by $(1/4)(3/4)^{n-2}$: $\frac{n-1}{2} \cdot \frac{1}{4} = \frac{9}{2} \cdot (3/4)$. $\frac{n-1}{8} = \frac{27}{8}$. $n-1 = 27$, so $n = 28$.
18. In which of the following scenarios can the random variable $X$ be correctly modelled by a binomial distribution?
    Answer: A student takes a 15-question multiple-choice test, where each question has 4 options. The student guesses every answer. $X$ is the number of correct answers. A binomial distribution requires four conditions: 1. A fixed number of trials ($n$). 2. Each trial must be independent. 3. Each trial has only two outcomes (success/failure). 4. The probability of success ($p$) is constant for each trial. A: This fits the model perfectly. $n=15$ is fixed, guessing is independent, each question is either correct (success) or incorrect (failure), and $p=1/4$ is constant. This is a binomial distribution. B: This is sampling without replacement from a small population. The probability of drawing a defective item changes after each draw, so the trials are not independent. This is a hypergeometric distribution. C: The number of trials is not fixed; it continues until a certain number of successes are achieved. This describes a negative binomial distribution. D: The weather on one day is often dependent on the weather of the previous day, so the independence assumption is likely violated.
19. A new drug is effective with a probability of 0.8. The drug is administered to a sequence of patients. What is the minimum number of patients that must be treated for the probability of at least one successful treatment to be greater than 0.999?
    Answer: 5 Let $X$ be the number of successful treatments among $n$ patients. Then $X \sim B(n, 0.8)$. We want to find the smallest integer $n$ such that $P(X \ge 1) > 0.999$. This is equivalent to $1 - P(X=0) > 0.999$, which simplifies to $P(X=0) < 0.001$. The probability of zero successes is $P(X=0) = \binom{n}{0} (0.8)^0 (0.2)^n = (0.2)^n$. We need to solve $(0.2)^n < 0.001$. We can test values of $n$: For $n=1$, $(0.2)^1 = 0.2$ (not < 0.001) For $n=2$, $(0.2)^2 = 0.04$ (not < 0.001) For $n=3$, $(0.2)^3 = 0.008$ (not < 0.001) For $n=4$, $(0.2)^4 = 0.0016$ (not < 0.001) For $n=5$, $(0.2)^5 = 0.00032$ (which is < 0.001) Therefore, the minimum number of patients is 5.

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