Question

1. $p(a)=\frac{1}{4},p(b)=\frac{1}{2},p(acup b)=?$ 12) $p(b)=\frac{1}{5},p(acup b)=\frac{7}{10},p(a)=?$

Explanation:

1. For the first - part (Question 10):

• Recall the formula for the probability of the union of two events: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). In the case where \(A\) and \(B\) are mutually - exclusive events (\(P(A\cap B) = 0\)), the formula simplifies to \(P(A\cup B)=P(A)+P(B)\).
• Given \(P(A)=\frac{1}{4}\) and \(P(B)=\frac{1}{2}\).
• # Explanation:
• ## Step 1: Apply the formula for the union of two events (assuming \(A\) and \(B\) are mutually - exclusive)
• We use the formula \(P(A\cup B)=P(A)+P(B)\).
• Substitute \(P(A)=\frac{1}{4}\) and \(P(B)=\frac{1}{2}\) into the formula. First, make the denominators the same. Since \(\frac{1}{2}=\frac{2}{4}\), then \(P(A\cup B)=\frac{1}{4}+\frac{2}{4}\).
• ## Step 2: Calculate the sum
• \(\frac{1 + 2}{4}=\frac{3}{4}\).
• # Answer:
• \(\frac{3}{4}\)

1. For the second - part (Question 12):

• Recall the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Again, assuming \(A\) and \(B\) are mutually - exclusive (\(P(A\cap B) = 0\)), we have \(P(A\cup B)=P(A)+P(B)\).
• Given \(P(B)=\frac{1}{5}\) and \(P(A\cup B)=\frac{7}{10}\).
• # Explanation:
• ## Step 1: Rearrange the formula for \(P(A)\)
• From \(P(A\cup B)=P(A)+P(B)\), we can solve for \(P(A)\) as \(P(A)=P(A\cup B)-P(B)\).
• ## Step 2: Make the denominators the same and subtract
• Since \(P(B)=\frac{1}{5}=\frac{2}{10}\), then \(P(A)=\frac{7}{10}-\frac{2}{10}\).
• \(P(A)=\frac{7 - 2}{10}=\frac{5}{10}=\frac{1}{2}\).
• # Answer:
• \(\frac{1}{2}\)

Answer:

1. For the first - part (Question 10):

• Recall the formula for the probability of the union of two events: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). In the case where \(A\) and \(B\) are mutually - exclusive events (\(P(A\cap B) = 0\)), the formula simplifies to \(P(A\cup B)=P(A)+P(B)\).
• Given \(P(A)=\frac{1}{4}\) and \(P(B)=\frac{1}{2}\).
• # Explanation:
• ## Step 1: Apply the formula for the union of two events (assuming \(A\) and \(B\) are mutually - exclusive)
• We use the formula \(P(A\cup B)=P(A)+P(B)\).
• Substitute \(P(A)=\frac{1}{4}\) and \(P(B)=\frac{1}{2}\) into the formula. First, make the denominators the same. Since \(\frac{1}{2}=\frac{2}{4}\), then \(P(A\cup B)=\frac{1}{4}+\frac{2}{4}\).
• ## Step 2: Calculate the sum
• \(\frac{1 + 2}{4}=\frac{3}{4}\).
• # Answer:
• \(\frac{3}{4}\)

1. For the second - part (Question 12):

• Recall the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Again, assuming \(A\) and \(B\) are mutually - exclusive (\(P(A\cap B) = 0\)), we have \(P(A\cup B)=P(A)+P(B)\).
• Given \(P(B)=\frac{1}{5}\) and \(P(A\cup B)=\frac{7}{10}\).
• # Explanation:
• ## Step 1: Rearrange the formula for \(P(A)\)
• From \(P(A\cup B)=P(A)+P(B)\), we can solve for \(P(A)\) as \(P(A)=P(A\cup B)-P(B)\).
• ## Step 2: Make the denominators the same and subtract
• Since \(P(B)=\frac{1}{5}=\frac{2}{10}\), then \(P(A)=\frac{7}{10}-\frac{2}{10}\).
• \(P(A)=\frac{7 - 2}{10}=\frac{5}{10}=\frac{1}{2}\).
• # Answer:
• \(\frac{1}{2}\)