In a simultaneous throw of two coins, the probability of getting at least one head is?

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

? 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

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

In a simultaneous throw of two dice
Simple Space = 6 x 6 = 36
Favorable cases are = (2, 6) (3, 5) (4, 4 ) (5, 3) (6, 2)
So, The required probability = 5/36

An ordinary year has 365 days i,e. 52 weeks and 1 day. So the probability that this day is a Sunday is 1/7.