Question

Three coins are tossed simultaneously. Find the probability of the following events(1).

Gettingexactly two heads.

(2) getting at least two heads

(3) getting no head

(4) getting at the most two tails​

Answer 1

possible outcome of tossing 3 coins are

(HHH),(TTT) , (HHT) , (TTH), (HTT) , (THH), (HTH) , (THT) = 8

(1) Getting exactly two heads:

E(Getting exactly two heads) = (HHT), (THH) , (HTH) = 3

$p(Getting \: exactly \: two \: heads) \: = \frac{3}{8}$

(2) getting at least two heads:

E(getting at least two heads) = (HHH) (THH) (HTH)

(HHT) = 4

$p(getting \: at \: least \: two \: heads) = \frac{4}{8} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2}$

(3) getting no head:

E(getting no head) = (TTT) = 1

$p(getting \: no \: head) = \frac{1}{8}$

(4) getting at the most two tails:

E(getting at the most two tails) = (THH) (HTH) (HHT) = 3

$p(getting \: at \: the \: most \: two \: tails) = \: \: \: \: \: \: \: \: \: \: \frac{3}{8}$