Question

A survey indicates that 41% of women in the United States consider reading their favorite leisure time activity. You randomly select four U.S. women and ask them if reading is their favorite leisure time activity. Find the probability that (1) exactly two of them respond yes, (2) at least two of them respond yes, and (3) fewer than two of them respond yes. (Source: Louis Harris & Associates)

Answer 1

1. The probability that exactly two of them respond yes is approximately 0.311.

2. The probability that at least two of them respond yes is approximately 0.569.

3. The probability that fewer than two of them respond yes is approximately 0.569.

Step-by-step explanation:

The probability of exactly two women responding yes can be calculated using the binomial probability formula, which is
`\( P(X=k) = \binom{n}{k} * p^k * (1-p)^(n-k) \), where \( n \) is the number of trials, \( k \)` is the number of successes, and \( p \) is the probability of success. In this case,
`\( n = 4 \), \( k = 2 \), and \( p = 0.41 \)`. Calculating this gives us the probability of exactly two women saying yes.

To find the probability that at least two of them respond yes, we sum the probabilities of exactly two, three, and four responding yes. This accounts for all scenarios where two or more women say yes. The probability of fewer than two responding yes is the complement of the probability of at least two responding yes. Therefore, subtracting the probability of at least two responding yes from 1 gives us the probability of fewer than two responding yes.

These probabilities provide insights into the likelihood of different outcomes when randomly selecting four U.S. women regarding their favorite leisure activity, given the survey data. The calculations are based on the assumption that the women's responses are independent and follow a binomial distribution.

Answer 2

(a) For the binomial distribution, assuming two possible outcomes per trial and independence.

(b) n = 4 (number of women asked), p = 0.41 (probability of a woman considering reading).

(c) Probability of exactly two women out of four considering reading: approximately 34.9%.

(a) For the binomial distribution, we assume that each trial (in this case, asking a woman if reading is her favorite leisure-time activity) has only two possible outcomes: success (a woman considers reading as her favorite leisure-time activity) or failure (a woman does not consider reading as her favorite leisure-time activity). Additionally, the trials are independent, meaning the outcome of one trial does not affect the outcome of another. Lastly, the probability of success (p) remains constant for each trial.

(b) In this scenario:

- n is the number of trials, which is the number of women asked, in this case, n = 4.

- p is the probability of success (a woman considering reading as her favorite leisure-time activity), which is p = 0.41 (or 41% as a decimal).

(c) Using the binomial probability formula:

`\[ P(X = k) = \binom{n}{k} * p^k * (1 - p)^(n - k) \]`

Where:

- P(X = k) is the probability of getting exactly k successes.

-
`\( \binom{n}{k} \)` is the number of ways to choose k successes out of n trials.

- p is the probability of success.

- 1 - p is the probability of failure.

For exactly two women responding "yes" (considering reading as their favorite leisure-time activity):

`\[ P(X = 2) = \binom{4}{2} * 0.41^2 * (1 - 0.41)^(4 - 2) \]`

First, calculate the combination:

`\[\binom{4}{2} = (4!)/(2!(4-2)!) = (4 * 3)/(2 * 1) = 6\]`

Now substitute this into the formula:

`\[P(X = 2) = 6 * 0.41^2 * (1 - 0.41)^(2)\]`

Calculate:

`\[P(X = 2) = 6 * 0.1681 * 0.3481 \approx 0.349\]`

Therefore, the probability that exactly two out of four randomly selected U.S. women consider reading as their favorite leisure-time activity is approximately 0.349 or 34.9%.