Question

events a and b are independent. find the missing probability.

1. ( p(a)=\frac{1}{4} ) ( p(b)=\frac{3}{5} ) ( p(b | a)=? )
2. ( p(b)=\frac{9}{20} ) ( p(a | b)=\frac{1}{5} ) ( p(a)=? )
3. ( p(a)=\frac{3}{10} ) ( p(b)=\frac{13}{20} ) ( p(a ) and ( b)=? )
4. ( p(b)=\frac{9}{20} ) ( p(a ) and ( b)=\frac{9}{100} ) ( p(a)=? )
5. ( p(a)=\frac{2}{5} ) ( p(a ) and ( b)=\frac{3}{10} ) ( p(operatorname{not} b)=? )
6. ( p(a)=\frac{7}{10} ) ( p(a ) or ( b)=\frac{173}{200} ) ( p(b)=? )

find the missing probability.

1. ( p(b)=\frac{2}{5} ) ( p(a ) and ( b)=\frac{1}{10} ) ( p(a | b)=? )
2. ( p(a)=\frac{3}{5} ) ( p(b | a)=\frac{3}{10} ) ( p(a ) and ( b)=? )
3. ( p(operatorname{not} a)=\frac{3}{5} ) ( p(a ) and ( b)=\frac{6}{25} ) ( p(b | a)=? )
4. ( p(b)=0.45 ) ( p(a ) or ( b)=0.72 ) ( p(b | a)=0.4 ) ( p(a)=? )

Explanation:

Step1: Use the formula for conditional probability

For independent events \(A\) and \(B\), \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). But since \(A\) and \(B\) are independent, \(P(A\cap B) = P(A)\times P(B)\). So \(P(B|A)=P(B)\).
Given \(P(B)=\frac{3}{5}\), so \(P(B|A)=\frac{3}{5}\)

Step2: Use the formula for conditional probability

For \(P(A|B)\), we know \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Given \(P(B)=\frac{9}{20}\) and \(P(A|B)=\frac{1}{5}\), then \(P(A\cap B)=P(A|B)\times P(B)=\frac{1}{5}\times\frac{9}{20}=\frac{9}{100}\). Since \(A\) and \(B\) are independent \(P(A\cap B) = P(A)\times P(B)\), so \(P(A)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{9}{100}}{\frac{9}{20}}=\frac{1}{5}\)

Step3: Use the formula for independent events

For \(P(A\cap B)\) when \(A\) and \(B\) are independent, \(P(A\cap B)=P(A)\times P(B)\). Given \(P(A)=\frac{3}{10}\) and \(P(B)=\frac{13}{20}\), then \(P(A\cap B)=\frac{3}{10}\times\frac{13}{20}=\frac{39}{200}\)

Step4: Use the formula for independent events

Since \(P(A\cap B) = P(A)\times P(B)\), given \(P(B)=\frac{9}{20}\) and \(P(A\cap B)=\frac{9}{100}\), then \(P(A)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{9}{100}}{\frac{9}{20}}=\frac{1}{5}\)

Step5: Use the formula for independent events

First, \(P(A\cap B) = P(A)\times P(B)\), so \(P(B)=\frac{P(A\cap B)}{P(A)}\). Given \(P(A)=\frac{2}{5}\) and \(P(A\cap B)=\frac{3}{10}\), then \(P(B)=\frac{\frac{3}{10}}{\frac{2}{5}}=\frac{3}{4}\). And \(P(\text{not }B)=1 - P(B)=1-\frac{3}{4}=\frac{1}{4}\)

Step6: Use the formula for \(P(A\cup B)\)

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Since \(A\) and \(B\) are independent \(P(A\cap B) = P(A)\times P(B)\). Let \(P(B)=x\). Given \(P(A)=\frac{7}{10}\) and \(P(A\cup B)=\frac{173}{200}\), then \(\frac{173}{200}=\frac{7}{10}+x-\frac{7}{10}x\).

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Step7: Use the formula for conditional probability

\(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Given \(P(B)=\frac{2}{5}\) and \(P(A\cap B)=\frac{1}{10}\), then \(P(A|B)=\frac{\frac{1}{10}}{\frac{2}{5}}=\frac{1}{4}\)

Step8: Use the formula for conditional probability

\(P(A\cap B)=P(B|A)\times P(A)\). Given \(P(A)=\frac{3}{5}\) and \(P(B|A)=\frac{3}{10}\), then \(P(A\cap B)=\frac{3}{10}\times\frac{3}{5}=\frac{9}{50}\)

Step9: First find \(P(A)\)

Since \(P(\text{not }A)=\frac{3}{5}\), then \(P(A)=1 - P(\text{not }A)=1-\frac{3}{5}=\frac{2}{5}\). Then use \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Given \(P(A\cap B)=\frac{6}{25}\), then \(P(B|A)=\frac{\frac{6}{25}}{\frac{2}{5}}=\frac{3}{5}\)

Step10: Use the formula for \(P(A\cup B)\)

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) and \(P(A\cap B) = P(B|A)\times P(A)\). Let \(P(A)=x\). Given \(P(B) = 0.45\), \(P(A\cup B)=0.72\) and \(P(B|A)=0.4\). Then \(P(A\cap B)=0.4x\).

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Answer:

1. \(\frac{3}{5}\)
2. \(\frac{1}{5}\)
3. \(\frac{39}{200}\)
4. \(\frac{1}{5}\)
5. \(\frac{1}{4}\)
6. \(\frac{11}{20}\)
7. \(\frac{1}{4}\)
8. \(\frac{9}{50}\)
9. \(\frac{3}{5}\)
10. \(0.45\)