# Three unbiased coins are tossed together. Find the probability of getting at least two heads.

Last updated date: 19th Mar 2023

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Hint : When we toss an unbiased coin we get either a head or tail. Here we are throwing three such coins. We are going to proceed with the thought of getting head and tail sequence on three coins.

Elementary events associated with the random experiment of tossing three coins are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.

Total number of elementary events = 8

If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event of “Getting at least two heads” occurs.

$\therefore $ Number of favorable elementary events = 4

$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$

Hence required probability = $\frac{4}{8} = \frac{1}{2}$

Note:

Probability of an event E is (Number of favorable outcomes to event E) / (Total number of possible outcomes in sample space)

$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$

In the given problem our event is getting at least two heads while tossing three coins.

Elementary events associated with the random experiment of tossing three coins are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.

Total number of elementary events = 8

If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event of “Getting at least two heads” occurs.

$\therefore $ Number of favorable elementary events = 4

$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$

Hence required probability = $\frac{4}{8} = \frac{1}{2}$

Note:

Probability of an event E is (Number of favorable outcomes to event E) / (Total number of possible outcomes in sample space)

$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$

In the given problem our event is getting at least two heads while tossing three coins.

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