Question

For the following exercises, four coins are tossed.

Find the probability of tossing either two heads or three heads.

Answer 1

Consider tossing of four coins:

The sample space for the given experiment is given by,

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

 

So, total number of possible outcomes is n(S) = 16.

 

Assume that E represents the event of tossing exactly two heads. Then,

E = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}

So, number of outcomes in the event E is n(E) = 6

 

Consider the formula for the probability of an event with equally likely outcomes,

P(E) = n(E)/n(S) ...... (1)

 

From formula (1) P(E) = n(E)/n(S), the probability of tossing exactly two heads is,

P(E) =  6/16

 

Assume that F represents the event of tossing exactly three heads. Then,

F = {THHH, HTHH, HHTH, HHHT}

 

So, number of outcomes in the event F is n(F) = 4

 

From formula (1) P(E) = n(E)/n(S), the probability of tossing exactly three heads is,

P(F) = 4/16

 

Events “tossing two heads” and “tossing three heads” are mutually exclusive events.

Use the formula for the probability of two mutually exclusive events,

P(E ∪ F) = P(E) + P(F) ...... (2)

 

The probability of tossing two heads or three heads is,

P(E ∪ F) = P(E) + P(F)

              = 6/16 + 4/16

              = 10/16

              = 5/8