? n(S) = (2)3 = 8
E = Event of getting 0, or 1 or 2 heads
= {TTT, TTH, THT, HTT, HHT, HTH, THH}
? n(E) = 7
? P(E) = n(E)/n(S) = 7/8
Total no. of balls = (6 + 8) = 14
No. of white balls = 8
? P(drawing a white ball) = 8/14 = 4/7
n(S) = Number of ways of drawing 2 balls out of 13 = 13C2 = 13 x 12 / 2= 78
n(E) = No. of ways of drawing 2 balls out of 5 = 5C2 = 5 x 4 / 2 = 10
? P(E) = n(E)/n(S) = 10/78 = 5/39
Number of cases favourable of E = 4
Total Number of cases = (5 + 4 ) = 9
? P(E) = 4/9
Clearly, n(S) = 20 and E = { 3, 6, 9, 12, 15, 18, 7, 14 } i.e., n(E) = 8
? P(E) = n(E)/ n(S) = 8/20 = 2/5
n(S) = 36
n(E) = {(5,6), (6,5)} = 2
? p(E) = n(E) / n(S) = 2/36 = 1/18
S= { HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} and
E = Event of getting exactly two heads = {HHT, HTH, THH}
? P(E) = n(E)/n(S) = 3/8
S = {HH, HT, TT, TH}
and E = {HH, HT, TH }
? P(E) = n(E)/n(S) = 3/4
Here S = {H, T}
and E = {H}
? P(E) = n(E)/n(S) = 1/2
The probability of absenting of the students in the class = 2/6 = 1/3
? The probability of missing his test = 1/5 x 1/3 = 1/15
In tossing a coin 2 times the sample space is 4 i,e (H, H), (H, T), (T, H), (T, T)
(1) If A1 denotes exactly one head
then A1 = {(H, T) (T, H) }
So, P(A1 ) = 2/4 = 1/2
(2) If A denotes at least one head
then A = {(H, T) (T, H) (H, H)}
? A = {(H, T) (T, H) (H, H )}
? P(A) = 3/4
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