Two coins are tossed once. Find the probability

Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.

Solution:

Total possible outcomes are . HH, HT, TT, TH, i.e., 4
(i) Favourable outcomes are HH, i.e., 1
So, P(2 heads)
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 1 }{ 4 }
(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail) = \\ \frac { 3 }{ 4 }

Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail
(iii) no tail
(iv) almost one tail.

Solution:

Two different coins are tossed simultaneously
Number of possible outcomes = (2)² = 4
Number of event having two tails = 1 i.e. (T, T)
(i) Probability of two tails will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 1 }{ 4 }
(ii) Number of events having one tail = 2 i.e. (TH) and (HT)
Probability of one tail will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 1 }{ 4 }
(iii) Number of events having no tail = 1 i.e. (HH)
Probability of having no tail will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 1 }{ 4 }
(iv) Atmost one tail
Number Of events having at the most one tail = 3 i.e. (TH), (HT, (TT)
Probability of at the most one tail will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 3 }{ 4 }

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