Question

1500 families were surveyed and following data was recorded about their maids at homes .types of maids, as only part time and only full time and both
frequency: 860 370 250 .a family is selected at random. Find the probability that the family selected has
1. Both types of maids
2. Part time maids
3. No maids.

Answer 1

Answer:

hey there is the answer

Answer 2

Answer:

The probability of the families selected has:

1) Both types of maids = $0.16$

2) Only part-time maids = $0.74$

3) No maids = $0.013$

Step-by-step explanation:

Given,

The total families, n(F) = $1500$

The type of maids:

• The part-time maids, n(A) =$860$
• The full-time maids, n(B) = $370$
• Both, n(A ∩ B) = $250$

1) The probability of the families selected having both types of maids:

Therefore,

• P(A ∩ B) = $\frac{n(A\cap B)}{n(F)}$    ------equation (a)

Here,

• P(A ∩ B) = The probability of families selected having both types of maids
• n(A ∩ B) = The frequency of both types of maids
• n(F) = The frequency of total families

After putting the given values in the equation (a), we get:

• P(A ∩ B) = $\frac{250}{1500}$ = $0.16$

2) The probability of the families selected having only part-time maids:

• P(A) = $\frac{n(A)+ n(A\cap B)}{n(F)}$    ------equation (b)

Here,

• P(A) = The probability of the families selected having only part-time maids
• n(A) = The frequency of part-time maids
• n(A ∩ B) = The frequency of both the types of maids
• n(F) = The frequency of the total families

After putting the given values in the equation (b), we get:

• P(A) = $\frac{860 + 250}{1500}$ = $0.74$

3) The probability of the families selected having no maids:

• Total families n(F) = $1500$
• Number of families that have maids = n(A) + n(B) + n(A ∩ B) = $860 + 370 + 250 = 1480$
• Number of families that don't have maids = $1500-1480 = 20$

Therefore,

• The probability of the families selected having no maids, P(N):
• P(N) = $\frac{20}{1500}$ = $0.013$