Question

in a company there are seven executives: four women and three men. three are selected at random to attend a management seminar. find these probabilities.
(a) all three selected will be women. round your answer to five decimal places.
the probability that all three people selected will be women is

Explanation:

Step1: Calculate the probability of selecting a woman first

The total number of executives is \(n = 7\), and the number of women is \(m=4\). The probability of selecting a woman first is \(P_1=\frac{4}{7}\).

Step2: Calculate the probability of selecting a woman second

After selecting one woman, the number of remaining executives is \(n_1 = 6\), and the number of remaining women is \(m_1 = 3\). The probability of selecting a woman second is \(P_2=\frac{3}{6}\).

Step3: Calculate the probability of selecting a woman third

After selecting two women, the number of remaining executives is \(n_2=5\), and the number of remaining women is \(m_2 = 2\). The probability of selecting a woman third is \(P_3=\frac{2}{5}\).

Step4: Calculate the combined probability

Since these are independent - sequential events (without replacement), the probability that all three are women is \(P = P_1\times P_2\times P_3\). Substitute the values: \(P=\frac{4}{7}\times\frac{3}{6}\times\frac{2}{5}=\frac{4\times3\times2}{7\times6\times5}=\frac{24}{210}\approx0.11429\)

Answer:

\(0.11429\)