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QUESTION 8



8.1. Given that:

$\begin{array}{l}

P(A) = 0,45 \\

P(B) = 0,3 \\

P(A \text{ or } B) = 0,615

\end{array}$



8.1.1. Calculate $P(A \text{ and } B)$. (2)



8.1.2. Are the events $A$ and $B$ independent? Show your calculations. (3)



8.2. The probability that the first answer in a Maths quiz competition will be correct is 0,4. If the first answer is correct, the probability of getting the next answer correct rises to...

Verified Answer :

  • Use the addition rule formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) to find \(P(A \cap B) = \boxed{0{,}135}\).
  • Evaluate independence by comparing \(P(A \cap B)\) with the product \(P(A) \times P(B) = 0{,}45 \times 0{,}3 = 0{,}135\).
  • Conclude that the events are independent since \(P(A \cap B) = P(A) \times P(B) = \boxed{\text{Yes}}\).

Explanation

  1. Calculate the probability of the intersection P(A and B)

To solve for the intersection of two events, \(A\) and \(B\), we use the Fundamental Addition Rule of Probability. This rule is defined as:
\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
Given our data, we have \(P(A) = 0{,}45\), \(P(B) = 0{,}3\), and \(P(A \cup B) = 0{,}615\). We can rearrange the formula to isolate the intersection probability \(P(A \cap B)\):
\[P(A \cap B) = P(A) + P(B) - P(A \cup B)\]
Substituting the known values:
\[P(A \cap B) = 0{,}45 + 0{,}3 - 0{,}615\]
\[P(A \cap B) = 0{,}75 - 0{,}615 = 0{,}135\]
Thus, the probability of both \(A\) and \(B\) occurring is \(0{,}135\).

  1. Check for independence between events A and B

To determine if the events \(A\) and \(B\) are independent, we check if the probability of both occurring is equal to the product of their individual probabilities. Mathematically, two events are independent if:
\[P(A \cap B) = P(A) \times P(B)\]
We already calculated \(P(A \cap B) = 0{,}135\). Now, we calculate the product of the individual probabilities:
\[P(A) \times P(B) = 0{,}45 \times 0{,}3\]
Performing the multiplication:
\[0{,}45 \times 0{,}3 = 0{,}135\]
Since \(P(A \cap B) = 0{,}135\) and \(P(A) \times P(B) = 0{,}135\), the condition for independence holds true. Therefore, events \(A\) and \(B\) are independent.

Examples

This concept is widely used in quality control, where you might check if two different types of machine faults are independent. For example, if the probability of a motor failing is \(0{,}05\) and the probability of a belt breaking is \(0{,}02\), knowing if these are independent helps engineers calculate the total probability of both happening simultaneously, which is \(\(0{,}05 \times 0{,}02 = 0{,}001\)\). This informs whether they need to design separate backup systems for each component.

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