A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads

#### Solution

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

n(S) = 8

Let A be the event of getting exactly two heads.

A = {HHT, HTH, THH}

n(A) = 3

P(A) = `("n"("A"))/("n"("S")) = 3/8`

Let B be the event of getting atleast one tail

B = {HHT, HTH, HTT, THH, THT, TTH, TTT}

n(B) = 7

P(B) = `("n"("B"))/("n"("S")) = 7/8`

Let C be the event of getting consecutively

C = {HHH, HHT, THH}

n(C) = 3

P(C) = `("n"("C"))/("n"("S")) = 3/8`

A ∩ B = {HHT, HTH, THH}

n(A ∩ B) = 3

p(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 3/8`

B ∩ C = {HHT, THH}

n(B ∩ C) = 2

P(B ∩ C) = `("n"("B" ∩ "C"))/("n"("S")) = 2/8`

A ∩ C = {HHT, THH}

n(A ∩ C) = 2

P(A ∩ C) = `("n"("A" ∩ "C"))/("n"("S")) = 2/8`

(A ∩ B ∩ C) = {HHT, THH}

n(A ∩ B ∩ C) = 2

P(A ∩ B ∩ C) = `("n"("A" ∩ "B" ∩ "C"))/("n"("S")) = 2/8`

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C)

= `3/8 + 7/8 + 3/8 - 3/8 - 2/8 - 2/8 + 2/8`

= `3/8 + 7/8 - 2/8`

= `(10 - 2)/8`

= `8/8`

= 1

The probability is 1.