The probability that a teacher will give one surprise test during any class meeting in a week is 1/5. If a student is absent twice. What is the probability that he will miss at least one test?

Here S = {H, T}
and E = {H}
? P(E) = n(E)/n(S) = 1/2

S = {HH, HT, TT, TH}
and E = {HH, HT, TH }
? P(E) = n(E)/n(S) = 3/4

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

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.

Clearly n(S) = 52 and there are 16 face cards.
? P(E) = 16/52 = 4/13

Clearly n(S) = 52. There are 26 red cards (including 2 kings ) and there are 2 more kings.
Let (E) be the event of getting either a red card or a king .
Then, n(E) = 28
? P(E) = n(E) / n(S) = 28/52 = 7/13