Exercises
Statistics I
Week 8 Graded Assignment

Week 8 Graded Assignment


Statistics Graded Assignment

1. Probability of the k-th Record Being the Last Five-Wicket Haul

Question: Zaheer Khan has taken $ m $ five-wicket hauls in his last $ n $ matches. His match records are selected at random, one by one, and analyzed. If none of the match records is analyzed more than once, then what is the probability that the $ k^{th} $ one analyzed is his last five-wicket haul match?

Options: a. $\frac{{ }^{m} C_{m-1} \times{ }^{n-m} C_{k-m}}{{ }^{n} C_{k-1}} \times \frac{1}{n-k+1}$ b. $\frac{{ }^{m} C_{m-1} \times{ }^{n-m} C_{k-m}}{{ }^{n} C_{k-1}}$ c. $\frac{1}{n-k+1}$ d. $\frac{{ }^{n-m} C_{k-m}}{{ }^{n} C_{k-1}}$

Answer: a

Solution: Let A be the event of getting exactly $(m-1)$ five-wicket haul matches when the first $(k-1)$ matches are analyzed. Let B be the event that the $k^{\text{th}}$ match analyzed is a five-wicket haul match.

$$ P(A) = \frac{{ }^{m} C_{m-1} \times{ }^{n-m} C_{k-m}}{{ }^{n} C_{k-1}} $$$$ P(B \mid A) = \frac{1}{n-k+1} $$$$ P(A \cap B) = P(A) \times P(B \mid A) = \frac{{ }^{m} C_{m-1} \times{ }^{n-m} C_{k-m}}{{ }^{n} C_{k-1}} \times \frac{1}{n-k+1} $$

Hence, option (a) is correct.

2. Probability of Correct Prediction Given Both Predict the Same Score

Question: A and B predict the outcomes of a cricket match and their chances of predicting the runs scored by a specific batsman correctly are $\frac{a}{b}$ and $\frac{c}{d}$ respectively, independent of each other. If the probability of them predicting the same wrong score is $\frac{p}{q}$, given that they predicted the same score, what is the probability that their answer is correct?

Options: a. $\frac{q-p}{q}$ b. $\frac{p(b-a)(d-c)}{q a c + p(b-a)(d-c)}$ c. $\frac{q a c}{q a c + p(b-a)(d-c)}$ d. $\frac{p(b-a)(d-c)}{q}$

Answer: c

Solution: Let:

  • $E_1$: Both A and B predicted the score correctly.
  • $E_2$: Exactly one predicted correctly.
  • $E_3$: Neither predicted correctly.
  • $E$: Both predicted the same score.
$$ P(E_1) = \frac{a}{b} \times \frac{c}{d} $$$$ P(E_2) = \frac{a}{b} \times \frac{d-c}{d} + \frac{b-a}{b} \times \frac{c}{d} $$$$ P(E_3) = \frac{b-a}{b} \times \frac{d-c}{d} $$$$ P(E \mid E_1) = 1, \quad P(E \mid E_2) = 0, \quad P(E \mid E_3) = \frac{p}{q} $$$$ P(E_1 \mid E) = \frac{P(E_1) P(E \mid E_1)}{P(E_1) P(E \mid E_1) + P(E_2) P(E \mid E_2) + P(E_3) P(E \mid E_3)} $$$$ = \frac{\frac{a c}{b d}}{\frac{a c}{b d} + \frac{(b-a)(d-c)}{b d} \times \frac{p}{q}} = \frac{q a c}{q a c + p(b-a)(d-c)} $$

Hence, option (c) is correct.

3. Probability that a Selected Item is Defective (from Three Factories)

Question: An item is produced in three factories $A, B$ and $C$. Factory $A$ produces $x$ times the number of items produced by factory $B$, and factories $B$ and $C$ produce the same number of items. It is known that $p%, q%, r%$ of the items produced by factories $A, B$ and $C$ respectively are defective. All items produced are stocked, and an item is selected at random. What is the probability that the selected item is defective?

Solution: Let the number of items produced by each of the factories $B$ and $C$ be $n$. Then, factory $A$ produces $x n$ items.

$$ P(A) = \sum_{i=1}^3 P(A \mid E_i) P(E_i) $$$$ = \frac{x}{x+2} \times \frac{p}{100} + \frac{1}{x+2} \times \frac{q}{100} + \frac{1}{x+2} \times \frac{r}{100} $$

For example, if $x=5, p=8, q=7, r=4$:

$$ P(A) = \frac{5}{7} \times \frac{8}{100} + \frac{1}{7} \times \frac{7}{100} + \frac{1}{7} \times \frac{4}{100} = \frac{51}{700} \approx 0.07 $$

4. Probability that a Defective Item is from Factory B

Question: If an item selected at random is found to be defective, what is the probability that it was produced by factory $B$?

Solution:

$$ P(E_2 \mid A) = \frac{P(E_2) P(A \mid E_2)}{P(A)} $$$$ = \frac{\frac{1}{x+2} \times \frac{q}{100}}{\frac{x}{x+2} \times \frac{p}{100} + \frac{1}{x+2} \times \frac{q}{100} + \frac{1}{x+2} \times \frac{r}{100}} $$

For example, $x=5, p=8, q=7, r=4$:

$$ P(E_2 \mid A) = \frac{\frac{1}{7} \times \frac{7}{100}}{\frac{51}{700}} = \frac{7}{51} \approx 0.14 $$

5. Probability that a Task is Completed (by Three Persons)

Question: A particular task is given to three persons, Manoj, Kalpana and Ananya, whose probabilities of completing it are $\frac{a}{b}, \frac{c}{d}$ and $\frac{e}{f}$ respectively, independent of each other. What is the probability that the task will be completed?

Solution:

$$ P(\text{task completed}) = 1 - \left(1 - \frac{a}{b}\right)\left(1 - \frac{c}{d}\right)\left(1 - \frac{e}{f}\right) $$

For example, $a=1, b=4, c=2, d=6, e=1, f=5$ (not given in solution, but similar logic applies):

$$ P(\text{task completed}) = 1 - \left(\frac{3}{4}\right)\left(\frac{4}{6}\right)\left(\frac{4}{5}\right) $$

But in the example provided in the solution:

$$ P(\text{task completed}) = 1 - \left(\frac{3}{4}\right)\left(\frac{4}{6}\right)\left(\frac{4}{5}\right) = 1 - \frac{48}{120} = 1 - 0.4 = 0.6 $$

(The example in the solution uses $P(A) = \frac{1}{4}, P(B) = \frac{2}{6}, P(C) = \frac{1}{5}$ and gets $1 - \frac{2}{5} = 0.6$.)

6. Value of $x$ Given $P(A^c) = m, P(B^c) = x, P(A \cup B) = n$

Question: If $A$ and $B$ are two independent events such that $P(A^c) = m$ and $P(B^c) = x$ and $P(A \cup B) = n$, then calculate the value of $x$.

Solution:

$$ n = 1 - m x $$$$ x = \frac{1 - n}{m} $$

For example, $m = 0.6, n = 0.7$:

$$ x = \frac{0.3}{0.6} = 0.5 $$

7–9. Intelligence Level and Researcher Data

Table Q8.1.G: Intelligence Level

ResearcherBelow Avg.Avg.Above Avg.
Xabc
Ydef

7. Probability that a Student is Below Average

Solution:

$$ P(\text{Below Avg.}) = \frac{a + d}{a + b + c + d + e + f} $$

For example, $a=76, b=70, c=54, d=50, e=23, f=27$:

$$ P = \frac{126}{300} = 0.42 $$

8. Probability that a Student is Average Given Researcher Y

Solution:

$$ P(\text{Avg.} \mid Y) = \frac{e}{d + e + f} $$

For example, $e=23, d+e+f=100$:

$$ P = \frac{23}{100} = 0.23 $$

9. Probability that Investigation is by Researcher X Given Below Average

Solution:

$$ P(X \mid \text{Below Avg.}) = \frac{a}{a + d} $$

For example, $a=76, d=50$:

$$ P = \frac{76}{126} \approx 0.603 $$

10. Probability that it Rained Given a Student Arrived Late

Question: During the monsoon, it rains one-third of the days and affects students’ travel to school. The probability that there will be heavy traffic on a rainy day is 0.5 and on a non-rainy day is 0.25. If it rains and there is heavy traffic, the probability of a student arriving late to school is 0.5. If it is a clear day and there is no traffic, this probability is reduced by $\frac{3}{8}$. In other possible situations, the probability of a student reaching school late is 0.25. If, on a randomly selected day, a student arrives late to school, then what is the probability that it rained that day?

Options: a. $\frac{1}{8}$ b. $\frac{6}{11}$ c. $\frac{11}{48}$ d. None of the above

Answer: b

Solution: Define:

  • $R$: Rainy day
  • $H$: Heavy traffic
  • $E$: Student is late
$$ P(R) = \frac{1}{3}, \quad P(H \mid R) = 0.5, \quad P(H \mid R^c) = 0.25 $$$$ P(E \mid R \cap H) = 0.5, \quad P(E \mid R^c \cap H^c) = 0.5 - \frac{3}{8} = \frac{1}{8}, \quad P(E \mid \text{other cases}) = 0.25 $$$$ P(E) = \frac{1}{12} + \frac{2}{48} + \frac{1}{24} + \frac{3}{48} = \frac{11}{48} $$$$ P(R \cap E) = \frac{1}{12} + \frac{1}{24} = \frac{3}{24} $$$$ P(R \mid E) = \frac{\frac{3}{24}}{\frac{11}{48}} = \frac{6}{11} $$

11. Value of $n$ Given Probabilities of Red T-Shirts

Question: There are two shops, $A$ and $B$, selling t-shirts in the market. Shop $A$ has stock of $n$ red and 2 black t-shirts and Shop $B$ has a stock of 2 red and $n$ black t-shirts. One of the shops is selected at random and two t-shirts are purchased from it. If both the t-shirts purchased are red and the probability that it was purchased from shop $A$ is $\frac{6}{7}$, find the value of $n$.

Options: a. 3 b. 4 c. 5 d. 6

Answer: b

Solution:

$$ P(A \mid E) = \frac{\frac{1}{2} \times \frac{n(n-1)}{(n+2)(n+1)}}{\frac{1}{2} \times \frac{n(n-1)}{(n+2)(n+1)} + \frac{1}{2} \times \frac{2}{(n+2)(n+1)}} = \frac{6}{7} $$$$ \frac{n(n-1)}{n(n-1) + 2} = \frac{6}{7} $$$$ 7n(n-1) = 6n(n-1) + 12 $$$$ n^2 - n - 12 = 0 $$$$ n = 4 $$

This covers all questions and solutions from the Week_8_Graded_-Solution.pdf1.

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  1. Week_8_Graded_-Solution.pdf ↩︎

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