Question

Three unbiased coins are tossed together. find the probability of getting (i) one head (ii) two heads (iii) all heads (iv) at least two heads.

Answer 1

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 event space,

S₁ = {TTH, THT, HTT}

Number of events, n(S₁) = 3

Thus, probability of getting one head

P(S₁) = $\frac{n(S_{1})}{n(S)}$

        = $\frac{3}{8}$

ii)

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

S₂ = {HHT, HTH, THH}

Number of events, n(S₂) = 3

Thus, probability of getting two heads

P(S₂) = $\frac{n(S_{2})}{n(S)}$

        = $\frac{3}{8}$

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) = $\frac{n(S_{4})}{n(S)}$

        = $\frac{4}{8}$

        = $\frac{1}{2}$

Answer 2

Answer: i) 3/8 ,  ii) 3/8 , iii) 1/2

Step-by-step explanation:  

Sample  space  of  tossing 3 coins together (Set  of total possible  events know  as  sample space)

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

n(S) = 8

i) One head

Events  having  one head,

E = {TTH , THT , HTT}

n(E) = 3

Hence,

Probability of  getting one head = n(E)/n(S) = 3/8

ii)Two heads

Events  having  two head,

E = {THH ,HHT , HTH}

n(E) = 3

Hence,

Probability of  getting one head = n(E)/n(S) = 3/8

(iii) All heads

Events  having  all heads,

E = {HHH}

n(E) = 1

Hence,

Probability of  getting one head = n(E)/n(S) = 1/8

(iv) at least two heads:- It means that at  heads two heads or  more that two as more as possible.

Events  having at least two heads,

E = {HHT, HHT , HTH , HHH}

n(E) = 4

Hence,

Probability of  getting one head = n(E)/n(S) = 4/8 = 1/2