Question

Three unbiased coins are tossed together. Find the probability of getting
(i) Two heads
(ii) At least two heads
(iii) No head

Answer 1

Number of possible outcomes in three tosses of coins- 2^3 = 8.

8 possibilities are- TTT, HHH, HTT, HHT, THH, TTH, THT, HTH.

i) 2 heads arise in 3 cases- HHT, THH, HTH.
Probability= 3/8

ii) At least 2 heads can arise in 4 cases- HHT, THH, HTH, HHH.
Probability= 4/8 = 1/2

iii)In 1 case no head can arise- TTT.
Probability= 1/8

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Answer 2

Answer :

Three unbiased coins are tossed together.

  Sample space, S

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

  Number of events, n(S) = 8

i)

For getting one head, we will have the sample space,

S₁ = {HHH, HHT, HTH, THH, TTH, THT, HTT}

Number of events, n(S₁) = 7

Thus, probability of getting one head

P(S₁) = $\frac{n(S_{1})}{n(S)}$

        = $\frac{7}{8}$

ii)

For getting two heads, we will have the sample space,

S₂ = {HHH, HHT, HTH, THH}

Number of events, n(S₂) = 4

Thus, probability of getting two heads

P(S₂) = $\frac{n(S_{2})}{n(S)}$

        = $\frac{4}{8}$

        = $\frac{1}{2}$

iii)

For getting all heads, we will have the sample space,

S₃ = {HHH}

Number of events, n(S₃) = 1

Thus, probability of getting all heads

P(S₃) = $\frac{n(S_{3})}{n(S)}$

        = $\frac{1}{8}$

iv)

For getting at least two heads, we will have the sample space,

S₄ = {HHH, HHT, HTH, THH}

Number of events, n(S₄) = 4

Thus, probability of at least two heads

P(S₄) = $\frac{n(S_{4})}{n(S)}$

        = $\frac{4}{8}$

        = $\frac{1}{2}$