Probability and Statistics Lecture 5

Continued

Probability and Statistics Lecture 4

Probability & Statistics Lecture 5

Question 1

Rahul travels by:

• Bus with probability $P(B)=\frac{2}{7}$
• Car with probability $P(C)=\frac{4}{7}$
• Motorcycle with probability $P(M)=\frac{1}{7}$

Given the probabilities of being late:

• $P(L|B)=\frac{3}{8}$
• $P(L|C)=\frac{1}{8}$
• $P(L|M)=\frac{4}{8}$

Find the probability of Rahul coming late $P(L)$.

Answer 1

The probability of Rahul coming late is given by:

$P(L)=P(B)\cdot P(L|B)+P(C)\cdot P(L|C)+P(M)\cdot P(L|M)$

Substituting the values:

$P(L)=\left(\frac{2}{7}\cdot \frac{3}{8}\right)+\left(\frac{4}{7}\cdot \frac{1}{8}\right)+\left(\frac{1}{7}\cdot \frac{4}{8}\right)$

Simplify:

$P(L)=\frac{6}{56}+\frac{4}{56}+\frac{4}{56}$

$P(L)=\frac{14}{56}=\frac{1}{4}$

Thus, the probability of Rahul coming late is $\frac{1}{4}$.

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Question 2

Question 2

Siya speaks either the truth or lies:

• Probability of truth: $P(T)=\frac{3}{8}$
• Probability of lying: $P(L)=\frac{5}{8}$

Part 1: Probability of her saying “6”

Let $P(6|T)$ be the probability of her saying “6” if she’s truthful (which is $\frac{1}{6}$ for a fair die), and $P(6|L)$ be the probability of her saying “6” if she’s lying (assumed as $1$ since she can fabricate it).

The total probability $P(6)$ is:

$P(6)=P(T)\cdot P(6|T)+P(L)\cdot P(6|L)$

Substituting the values:

$P(6)=\left(\frac{3}{8}\cdot \frac{1}{6}\right)+\left(\frac{5}{8}\cdot 1\right)$

$P(6)=\frac{3}{48}+\frac{40}{48}=\frac{43}{48}$

Thus, the probability of Siya saying “6” is $\frac{43}{48}$.

Part 2: Probability of truth given she says “6”

We use Bayes’ theorem:

$P(T|6)=\frac{P(T)\cdot P(6|T)}{P(6)}$

Substituting the values:

$P(T|6)=\frac{\frac{3}{8}\cdot \frac{1}{6}}{\frac{43}{48}}$

$P(T|6)=\frac{\frac{3}{48}}{\frac{43}{48}}=\frac{3}{43}$

Thus, the probability of Siya telling the truth while saying “6” is $\frac{3}{43}$.

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Question 3

There are always 50% spam mails:

• $P(S)=0.5$, $P(\neg S)=0.5$
• Spam filter detects 99% of spam: $P(D|S)=0.99$
• False positive rate: 5% of non-spam: $P(D|\neg S)=0.05$

Find the probability of a non-spam mail being falsely detected as spam ($P(\neg S|D)$).

We use Bayes’ theorem:

$P(\neg S|D)=\frac{P(\neg S)\cdot P(D|\neg S)}{P(D)}$

Where $P(D)$ is the total probability of detection:

$P(D)=P(S)\cdot P(D|S)+P(\neg S)\cdot P(D|\neg S)$

Substitute the values:

$P(D)=(0.5\cdot 0.99)+(0.5\cdot 0.05)$ $P(D)=0.495+0.025=0.52$

Now, calculate $P(\neg S|D)$:

$P(\neg S|D)=\frac{0.5\cdot 0.05}{0.52}$ $P(\neg S|D)=\frac{0.025}{0.52}\approx 0.0481$

Thus, the probability of a non-spam mail being falsely detected as spam is approximately $0.0481$ or $4.81\%$.

References

Continued

Probability and Statistics Lecture 6

N/A

Information

• Date: 2025.01.15
• Time: 14:05