Total number of families S = 1500
n(S) = 1569
Let P be the event of selecting a family having part time maids and F be the event of selecting a family having full time maids.
(i) Both types of maids Let P \(\cap\) F be the event of selecting a family having both types of maids.
Let (P \(\cap\) F) = 259
\(\mathrm{P}(\mathrm{P} \cap \mathrm{F})=\frac{\mathrm{n}(\mathrm{P} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{250}{1500}=\frac{1}{6}\)
(ii) Part time maids Part time maids = only part time maid + both
= 860 + 250 = 1110 '
n(P) = 1110
\(\mathrm{p}(\mathrm{P})=\frac{\mathrm{n}(\mathrm{P})}{\mathrm{n}(\mathrm{S})}=\frac{1110}{1500}=\frac{111}{150}\)
(iii) No maids Let (P \(\cup\) P)'be the event of choosing a family not having maids and (P u F) be the event of choosing a family having part time or fuli time or both maids.
n(P \(\cup\) F ) = only n(P) + only n(F) + n(p \(\cap\) p) = 850 + 370 + 250
n(P\(\cup\)F) = 1480
n(P\(\cup\)F)' = n(S)-n(P\(\cup\)F)
= 1500 - 1480
n(P \(\cup\) F)' = 20
\(\mathrm{P}(\mathrm{P} \cup \mathrm{F})^{\prime}=\frac{\mathrm{n}(\mathrm{P} \cup \mathrm{F})^{\prime}}{\mathrm{n}(\mathrm{S})}=\frac{20}{1500}=\frac{1}{75}\)

No. of youngsters = 400
n(S)
No. of youngsters having voter id = 191
n(A)
No. of youngsters do not have their voter id
n(B) = 400 - 191 = 209
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 209 }{ 400 } \).

Total n(S) = 10,000
Defective n(A) = 25
Number of defective laptop is 25
Number of good laptop is = 10000 - 25 = 9975
Let A be the event of choosing a good laptop. n(A) = 9975
\(\mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{9975}{10000}\)
P(A) = 0.9975
Probability of getting a good laptop is 0.9975

(i) What is the probability that the spinner will not land on a multiple of 2?
(ii) What is the probability that the spinner will land on an odd number?

When two dice are rolled
Sample space
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4),(4,5),(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
n(S) = 36
(i) Event of the sum is equal to 1 = 0
∴ Probability = \(\frac { 0 }{ n(S) } \) = 0
(ii) Event of the sum is equal to 4
B = {(1, 3), (2, 2), (3, 1)}
n(B) = 3
P(B)= \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 12 } \)
(iii) Event of the sum is equal to less than 13
C= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
n(C) = 36
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 6 } \).