18. a) the masinde muliro university uses three buses a, b and c. on any day the probability of the buses operating are 0.75, 0.4 and 0.5 respectively. use a tree diagram to find the probability that on a given day
i) all the buses are not operating (3 marks)
ii) only one bus is operating (2 marks)
iii) at least one bus is operating (2 marks)
asket a contains 5 oranges and 3 mangoes while basket b contains 4 oranges and 3 mangoes. a basket is
d randomly and two fruits drawn from it, one at a time without replacement. find the probability that the
picked are of the same type. (3 marks)

Question

18. a) the masinde muliro university uses three buses a, b and c. on any day the probability of the buses operating are 0.75, 0.4 and 0.5 respectively. use a tree diagram to find the probability that on a given day
 i) all the buses are not operating (3 marks)
 ii) only one bus is operating (2 marks)
 iii) at least one bus is operating (2 marks)
asket a contains 5 oranges and 3 mangoes while basket b contains 4 oranges and 3 mangoes. a basket is
d randomly and two fruits drawn from it, one at a time without replacement. find the probability that the
 picked are of the same type. (3 marks)

Accepted Answer

### i) All the buses are not operating
# Explanation:
## Step 1: Find the probability that each bus is not operating.
The probability that bus A is not operating is $1 - 0.75 = 0.25$.
The probability that bus B is not operating is $1 - 0.4 = 0.6$.
The probability that bus C is not operating is $1 - 0.5 = 0.5$.

## Step 2: Multiply the probabilities (since the events are independent).
The probability that all three buses are not operating is $0.25\times0.6\times0.5$.
$$
0.25\times0.6 = 0.15 \
0.15\times0.5 = 0.075
$$

# Answer:
$0.075$

### ii) Only one bus is operating
# Explanation:
## Step 1: Identify the three cases (only A, only B, only C operating).
- **Case 1: Only A operates**
  Probability A operates: $0.75$, B does not: $0.6$, C does not: $0.5$.
  Probability: $0.75\times0.6\times0.5$.

- **Case 2: Only B operates**
  Probability A does not: $0.25$, B operates: $0.4$, C does not: $0.5$.
  Probability: $0.25\times0.4\times0.5$.

- **Case 3: Only C operates**
  Probability A does not: $0.25$, B does not: $0.6$, C operates: $0.5$.
  Probability: $0.25\times0.6\times0.5$.

## Step 2: Calculate each case and sum them.
- Case 1: $0.75\times0.6\times0.5 = 0.225$
- Case 2: $0.25\times0.4\times0.5 = 0.05$
- Case 3: $0.25\times0.6\times0.5 = 0.075$

Sum: $0.225 + 0.05 + 0.075 = 0.35$

# Answer:
$0.35$

### iii) At least one bus is operating
# Explanation:
## Step 1: Use the complement rule (1 - probability all are not operating).
From part (i), the probability all buses are not operating is $0.075$.

## Step 2: Subtract from 1.
The probability that at least one bus is operating is $1 - 0.075$.
$$
1 - 0.075 = 0.925
$$

# Answer:
$0.925$

### Fruit Basket Problem (Probability same type)
# Explanation:
## Step 1: Probability of choosing basket A.
There are 2 baskets, so probability of choosing A is $\frac{1}{2}$.

## Step 2: Probability of same type from basket A.
Basket A has 5 oranges ($O$) and 3 mangoes ($M$), total $8$ fruits.
- Probability two oranges: $\frac{5}{8}\times\frac{4}{7}$ (without replacement).
- Probability two mangoes: $\frac{3}{8}\times\frac{2}{7}$.

Sum for basket A: $\frac{5}{8}\times\frac{4}{7} + \frac{3}{8}\times\frac{2}{7}$.

## Step 3: Probability of choosing basket B.
Probability of choosing B is $\frac{1}{2}$.

## Step 4: Probability of same type from basket B.
Basket B has 4 oranges and 3 mangoes, total $7$ fruits.
- Probability two oranges: $\frac{4}{7}\times\frac{3}{6}$.
- Probability two mangoes: $\frac{3}{7}\times\frac{2}{6}$.

Sum for basket B: $\frac{4}{7}\times\frac{3}{6} + \frac{3}{7}\times\frac{2}{6}$.

## Step 5: Combine the probabilities (basket A or basket B).
Total probability = (Probability choose A $\times$ Probability same type in A) + (Probability choose B $\times$ Probability same type in B).

### Calculations:
- **Basket A same type:**
  $\frac{5}{8}\times\frac{4}{7} = \frac{20}{56}$, $\frac{3}{8}\times\frac{2}{7} = \frac{6}{56}$.
  Sum: $\frac{20 + 6}{56} = \frac{26}{56} = \frac{13}{28}$.

- **Basket B same type:**
  $\frac{4}{7}\times\frac{3}{6} = \frac{12}{42}$, $\frac{3}{7}\times\frac{2}{6} = \frac{6}{42}$.
  Sum: $\frac{12 + 6}{42} = \frac{18}{42} = \frac{3}{7}$.

- **Total probability:**
  $\frac{1}{2}\times\frac{13}{28} + \frac{1}{2}\times\frac{3}{7}$.
  Convert $\frac{3}{7}$ to $\frac{12}{28}$:
  $\frac{13}{56} + \frac{12}{56} = \frac{25}{56} \approx 0.446$.

# Answer:
$\frac{25}{56}$ (or approximately $0.446$)

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QUESTION IMAGE

Question

Explanation:

i) All the buses are not operating

Step 1: Find the probability that each bus is not operating.

Step 2: Multiply the probabilities (since the events are independent).

Step 1: Identify the three cases (only A, only B, only C operating).

Step 2: Calculate each case and sum them.

Step 1: Use the complement rule (1 - probability all are not operating).

Step 2: Subtract from 1.

Answer:

ii) Only one bus is operating