Question

Three different coins are tossed together. find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails

Answer 1

(i) P(Getting exactly two heads) = 3 / 8

(ii) P(Getting at least two heads) = 1 / 2

(iii) P(Getting at least two heads) = 1 / 2

Step-by-step explanation:

Possible outcomes when three coins are tossed together:

= 2 * 2 * 2 = 8 Possible Outcomes

Always keep in mind this formula while calculating Probability of an event:

$\fbox{P(E) = No Of Outcomes / Total No Of Outcomes}$

(i) No of possible outcomes for getting exactly two heads = HHT, HTH and THH.

3 possible outcomes for getting exactly two heads.

P(Getting exactly two heads) = 3 / 8

(ii) No of possible outcomes for getting at least two heads = HHT, HTH, THH and HHH.

4 possible outcomes for getting at least two heads.

P(Getting at least two heads) = 4 / 8

= 2 / 4

= 1 / 2

(iii) No of possible outcomes to get at least two tails = TTH, THT, HTT and TTT.

4 possible outcomes to get at least two tails.

P(Getting at least two tails) = 4 / 8

= 2 / 4

= 1 / 2

Answer 2

Answer:

Total Possible Outcomes= 8

(i) P(Getting exactly two heads) = 3 / 8

[1,2]  [1,3]  [2,3]

(ii) P(Getting at least two heads) = 1 / 2

[1,2]  [1,3]  [2,3]  [1,2,3]

(iii) P(Getting at least two tails) = 1 / 2

[1,2]  [1,3]  [2,3]  [1,2,3]