Question

2.2 for two events a and b, it is given that:

• p(a)=0.3
• p(a and b)=0.2
• p(not b)=0.25

where a and b are two different events.
2.2.1 draw a venn - diagram to represent this information.
2.2.2 determine p(a or b).
2.3 sizwe has enrolled with a driving school that guarantees that he will pass the test to obtain his drivers licence. the probability that he will pass the driving on his first attempt is 2/5. if he fails on his first attempt, the probability that sizwe will pass the test on any attempt in the future is 2/3. calculate the probability that sizwe will pass the test in:
2.3.1 two attempts
2.3.2 three attempts
2.3.3 four or more attempts

Explanation:

Step1: Recall the formula for $P(A\ or\ B)$

$P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)$
First, find $P(B)$. We know that $P(\text{not }B) = 0.25$, so $P(B)=1 - P(\text{not }B)=1 - 0.25 = 0.75$.

Step2: Substitute values into the formula

$P(A)=0.3$, $P(B)=0.75$ and $P(A\ and\ B)=0.2$.
$P(A\ or\ B)=0.3 + 0.75- 0.2$
$P(A\ or\ B)=0.85$

Step3: Calculate probability of passing in two attempts

The probability of failing the first - attempt and passing the second attempt. The probability of failing the first attempt is $1-\frac{2}{5}=\frac{3}{5}$, and the probability of passing the second attempt given failure on the first is $\frac{2}{3}$. So the probability of passing in two attempts is $\frac{3}{5}\times\frac{2}{3}=\frac{2}{5}$.

Step4: Calculate probability of passing in three attempts

The probability of failing the first two attempts and passing the third attempt. The probability of failing the first attempt is $\frac{3}{5}$, the probability of failing the second attempt given failure on the first is $1 - \frac{2}{3}=\frac{1}{3}$, and the probability of passing the third attempt given failure on the first two is $\frac{2}{3}$. So the probability is $\frac{3}{5}\times\frac{1}{3}\times\frac{2}{3}=\frac{2}{15}$.

Step5: Calculate probability of passing in four or more attempts

First, find the probability of passing in one, two or three attempts.
Probability of passing in one attempt is $\frac{2}{5}$, in two attempts is $\frac{2}{5}$, and in three attempts is $\frac{2}{15}$.
The sum of these probabilities is $\frac{2}{5}+\frac{2}{5}+\frac{2}{15}=\frac{6 + 6+ 2}{15}=\frac{14}{15}$.
The probability of passing in four or more attempts is $1-\frac{14}{15}=\frac{1}{15}$.

Answer:

2.2.1: (Draw a Venn - diagram with two overlapping circles labeled A and B. The intersection part is labeled with 0.2, the non - intersection part of A is $0.3 - 0.2=0.1$, and the non - intersection part of B is $0.75 - 0.2 = 0.55$)
2.2.2: $0.85$
2.3.1: $\frac{2}{5}$
2.3.2: $\frac{2}{15}$
2.3.3: $\frac{1}{15}$