A salesman has 60% chance of making a sale to each customer. The behaviour of successive customers is independent. If two customers A and B enter the shop, what is the probability... A salesman has 60% chance of making a sale to each customer. The behaviour of successive customers is independent. If two customers A and B enter the shop, what is the probability that the salesman will make a sale to A or B? | Probability that a man will be alive 25 years is 0.3 and the probability that his wife will be alive 25 years hence is 0.4. Find the probability that (i) both will be alive (ii) only the man will be alive (iii) only the woman will be alive (iv) none will be alive (v) at least one of them will be alive. | If P(A) = 2P(B) = P(A/B) = 0.4, find: (i) Both A and B occurs (ii) Only B occurs (iii) At least one occur (iv) None occurs (neither A and B occurs). | The incidence of occupational disease in an industry is such that the workers have 20% chances of suffering from it. What is the probability that one of six workers, 4 or more will contact the disease? | The probability of failure in physics practical examination is 20%. If 25 batches of 6 students each take the examination, in how many batches 4 or more students would pass? | It is known from past experience that in a certain plant, there are, on average, 4 industrial accidents per year. Find the probability that in a given year, there will be less than 4 accidents. Assume Poisson distribution. | Between the hours of 2 and 4 p.m., the average number of phone calls per minute coming into the switchboard of a company is 2. Find the probability that during that particular minute there will be no phone call at all. | A random variable X has the following probability function: Value of X: -2, -1, 0, 1, 2, 3. Probability: 0.1, k, 0.2, 2k, 0.3, k. Find the value of k, and calculate mean and variance of X. Read more

Steps to Solve

1. Calculating Probability of Sale to A or B

To find the probability of making a sale to either customer A or B, we can use the complementary probability. The probability that the salesman does not make a sale to a customer is $1 - 0.6 = 0.4$.

The probability that he does not make a sale to either A or B (both failing) is: $$ P(\text{not A and not B}) = P(\text{not A}) \times P(\text{not B}) = 0.4 \times 0.4 = 0.16 $$

Thus, the probability that he makes at least one sale to either A or B is: $$ P(\text{A or B}) = 1 - P(\text{not A and not B}) = 1 - 0.16 = 0.84 $$

2. Calculating Probability for Man and Wife

Given:

• Probability that the man is alive, $P(M) = 0.3$
• Probability that the woman is alive, $P(W) = 0.4$

(i) Both will be alive: $$ P(M \cap W) = P(M) \times P(W) = 0.3 \times 0.4 = 0.12 $$

(ii) Only the man will be alive: $$ P(M \cap \text{not } W) = P(M) \times (1 - P(W)) = 0.3 \times 0.6 = 0.18 $$

(iii) Only the woman will be alive: $$ P(\text{not } M \cap W) = (1 - P(M)) \times P(W) = 0.7 \times 0.4 = 0.28 $$

(iv) None will be alive: $$ P(\text{not } M \cap \text{not } W) = (1 - P(M)) \times (1 - P(W)) = 0.7 \times 0.6 = 0.42 $$

(v) At least one will be alive: $$ P(M \cup W) = 1 - P(\text{not } M \cap \text{not } W) = 1 - 0.42 = 0.58 $$

3. Finding Probabilities for Events A and B

Given $P(A) = 2P(B) = P(A|B) = 0.4$, we can first find $P(B)$: Let $P(B) = x$, then $P(A) = 2x$ and using the conditional probability: $$ P(A|B) = \frac{P(A \cap B)}{P(B)} = 0.4 $$

From $P(A \cap B) = P(A|B) \times P(B) = 0.4 \times x$. Thus we have: $$ 2x \times x = 0.4x $$

This means: $$ 2x^2 = 0.4x \Rightarrow 2x^2 - 0.4x = 0 \Rightarrow x(2x - 0.4) = 0 $$

So $x = 0$ or $x = 0.2$.

Substituting $x = 0.2$ gives $P(A) = 0.4$.

(i) Both A and B occur: $$ P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.2 = 0.08 $$

(ii) Only B occurs: $$ P(\text{not } A \cap B) = P(B) - P(A \cap B) = 0.2 - 0.08 = 0.12 $$

(iii) At least one occurs: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.2 - 0.08 = 0.52 $$

(iv) None occurs: $$ P(\text{not } A \cap \text{not } B) = 1 - P(A \cup B) = 1 - 0.52 = 0.48 $$

4. Occupational Disease Probability

Each worker has a 20% chance of getting the disease ($p = 0.2$). For 6 workers, we are interested in $P(X \geq 4)$ where $X$ is the number of workers getting the disease.

We can use the binomial distribution: $$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$

Calculating $P(X \geq 4)$: $$ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) $$

Calculating each:

• For $k = 4$: $$ P(X = 4) = \binom{6}{4} (0.2)^4 (0.8)^2 = 15 \cdot 0.0016 \cdot 0.64 = 0.1536 $$

• For $k = 5$: $$ P(X = 5) = \binom{6}{5} (0.2)^5 (0.8)^1 = 6 \cdot 0.00032 \cdot 0.8 = 0.001536 $$

• For $k = 6$: $$ P(X = 6) = \binom{6}{6} (0.2)^6 (0.8)^0 = 0.000064 $$

Now sum these: $$ P(X \geq 4) = 0.1536 + 0.001536 + 0.000064 \approx 0.1552 $$

5. Passing Physics Examination

The probability of passing = $1 - 0.2 = 0.8$.

Using the binomial distribution for 25 batches (n=25, p=0.8): We want the expected number of batches where 4 or more pass. Using $P(Y \geq 4)$: The expected number of successful batches $= 25 \cdot P(Y \geq 4)$.

6. Poisson Distribution for Accidents

Using Poisson distribution with $\lambda = 4$:

The probability of less than 4 accidents can be calculated by summing $P(X < 4)$: $$ P(X < k) = e^{-\lambda} \sum_{x=0}^{k-1} \frac{\lambda^x}{x!} $$

With $\lambda = 4$, for $k = 4$: $$ P(X < 4) = e^{-4} (\frac{4^0}{0!} + \frac{4^1}{1!} + \frac{4^2}{2!} + \frac{4^3}{3!}) $$

Calculating this gives an approximate probability.

7. Phone Calls Probability

Given rate = 2 calls per minute, using Poisson with $\lambda = 2$ for zero calls: $$ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-2} $$

8. Finding k for Random Variable X

Given:

• Values of $X$: -2, -1, 0, 1, 2, 3
• Probabilities: 0.1, $k$, 0.2, $2k$, 0.3, $k$

The total probability must equal 1: $$ 0.1 + k + 0.2 + 2k + 0.3 + k = 1 $$

This yields: $$ 0.6 + 4k = 1 \Rightarrow 4k = 0.4 \Rightarrow k = 0.1 $$

Mean and variance can be calculated once probability check is satisfied.