(i) ‘8’
(ii) an even number
(iii) a perfect square
(iv) a negative number
(v) a number less than 13
Since, Sample Space or
S=1,2,3,4,5,6,7,8,9,10,11,12
Then,
(i) ‘8’
Solution:
S=1,2,3,4,5,6,7,8,9,10,11,12
Let A be the event selecting ‘8’, then n(A)=8
P(A)=(n(A))/(n(S))
P(A)=1/12
As 8 numbered card is only one in the box, so the Probability of drawing a
card ‘8’ from box will be 1/12
(ii) an even number
Solution:
S=1,2,3,4,5,6,7,8,9,10,11,12
Let A be the event selecting ‘even number’, then n(A)=2,4,6,8,10,12
P(A)=(n(A))/(n(S))
P(A)=6/12
P(A)=1/2
As Even number cards is six in the box, so the Probability of drawing a card an even number from box will be 1/2
(iii) a perfect square
Solution:
S=1,2,3,4,5,6,7,8,9,10,11,12
Let A be the event selecting ‘a perfect square’, then n(A)=1, 4, 9
P(A)=(n(A))/(n(S))
P(A)=3/12
P(A)=1/4
As a perfect square number cards is 3 in the box, so the Probability of drawing a card a
perfect square from box will be 1/4
(iv) a negative number
Solution:
S=1,2,3,4,5,6,7,8,9,10,11,12
Let A be the event selecting ‘a negative number’, then n(A)=0
P(A)=(n(A))/(n(S))
P(A)=0/12
P(A)=0
As there is no any negative number card in the box, so the Probability of drawing a card a negative number from box will be 0
(v) a number less than 13
Solution:
S=1,2,3,4,5,6,7,8,9,10,11,12
Let A be the event selecting ‘a number less than 13’, then n(A)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
P(A)=(n(A))/(n(S))
P(A)= 12/12
P(A)= 1
As all numbers in the box are less than 13, so the Probability of drawing a card a
Number less than 13 from box will be 1.