**The probability of getting at most two heads when three coins are tossed simultaneously.**

The **Answer to the Question**

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**Here's the Solution to this Question**

we define the sample space as S = { HHH, HHT, HTH, HTT, TTT, TTH, THT, THH }

At most two heads, implies either 0, 1 or exactly two heads.

The number of combinations with at most two heads are:

HHT, HTH, HTT, TTH, THT, THH, TTT

Therefore, the probability of getting at most two heads =( (1/8) +(1/8) + (1/8) + (1/8) + (1/8) + (1/8) + (1/8) ) which equals to (7/8)