1500 families were surveyed and following date was recorded about their maids at home
`{:(" Type of maids "," Only part time "," Only full time "," Both "),(" Number of families"," 860"," 370"," 250"):}`
A familiy is selected at random. Find the probability that the family selected has
(i) Both types of maids (ii) Part time maids (iii) No maids

Explanation

`{:(" n(s)=1500"" (Total families)"),(" n(A)=860"" (Part time maids)"),(" n(B)=370"" (Only full time)"),(n(A cap B)=250" " (Both)"):}`
(i) `p(AcapB) = (n(AcapB))/(n(S))=(cancel(250))/(underset(6)cancel(1500))=(1)/(6)`
(ii) `P(A) =(n(A))/(n(S)) = (cancel(860))/(cancel(1500))=(43)/(75)`
(iii) Total families n(S) = 1500
No. of families have maids = 860 + 250 + 370 = 1480
No. of families do not have maids = 1500 - 1480
n(A) = 20
`:. P(A) = (n(A))/(n(S)) = (cancel(20))/(cancel(1500)) = (1)/(75)`