Question

elias writes the numbers 1 through 20 on separate slips of paper. there are 16 white slips of paper and four yellow slips of paper. there are eight odd numbers on white slips, and the rest of the odd numbers are on yellow slips. are the events odd and yellow independent? no, because the probability of choosing a yellow slip is not equal to the probability of choosing a yellow slip given an odd number yes, because the probability of choosing an odd number on a yellow slip of paper is equal to the probability of choosing an odd number no, because the probability of choosing an odd number is not equal to the probability of choosing an odd number on a yellow slip yes, because the probability of choosing an odd

Explanation:

Step1: Calculate total number of slips

Total slips \(n = 20\)

Step2: Calculate number of odd numbers

Numbers from \(1\) to \(20\), odd numbers \(n(\text{odd})=10\)

Step3: Calculate number of yellow slips

Yellow slips \(n(\text{yellow}) = 4\)

Step4: Calculate number of odd - yellow slips

Since there are \(8\) odd numbers on white slips, odd - yellow slips \(n(\text{odd}\cap\text{yellow})=10 - 8=2\)

Step5: Calculate \(P(\text{odd})\)

\(P(\text{odd})=\frac{n(\text{odd})}{n}=\frac{10}{20}=\frac{1}{2}\)

Step6: Calculate \(P(\text{yellow})\)

\(P(\text{yellow})=\frac{n(\text{yellow})}{n}=\frac{4}{20}=\frac{1}{5}\)

Step7: Calculate \(P(\text{odd}\cap\text{yellow})\)

\(P(\text{odd}\cap\text{yellow})=\frac{n(\text{odd}\cap\text{yellow})}{n}=\frac{2}{20}=\frac{1}{10}\)

Step8: Calculate \(P(\text{odd})\times P(\text{yellow})\)

\(P(\text{odd})\times P(\text{yellow})=\frac{1}{2}\times\frac{1}{5}=\frac{1}{10}\)

Step9: Calculate \(P(\text{yellow}|\text{odd})\)

By formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\), so \(P(\text{yellow}|\text{odd})=\frac{P(\text{odd}\cap\text{yellow})}{P(\text{odd})}=\frac{\frac{1}{10}}{\frac{1}{2}}=\frac{1}{5}\)

Since \(P(\text{odd}\cap\text{yellow}) = P(\text{odd})\times P(\text{yellow})\) and \(P(\text{yellow}|\text{odd})=P(\text{yellow})\), the events are not independent. Because for independent events \(P(A\cap B)=P(A)\times P(B)\) and \(P(A|B) = P(A)\) (or \(P(B|A)=P(B)\)). Here \(P(\text{yellow}|\text{odd})=\frac{1}{5}\), \(P(\text{yellow})=\frac{1}{5}\) is wrong. The correct check is \(P(\text{odd}|\text{yellow})=\frac{P(\text{odd}\cap\text{yellow})}{P(\text{yellow})}=\frac{\frac{1}{10}}{\frac{1}{5}}=\frac{1}{2}\) and \(P(\text{odd})=\frac{1}{2}\). But another way: \(P(\text{yellow})=\frac{4}{20}\), \(P(\text{yellow}|\text{odd})=\frac{2}{10}\). Since \(\frac{4}{20}
eq\frac{2}{10}\)

Answer:

no, because the probability of choosing a yellow slip is not equal to the probability of choosing a yellow slip given an odd number