Question

Two unbiased coins are tossed. What is the probability of getting at least two heads?
A) $\dfrac{1}{2}$
B) $\dfrac{1}{4}$
C) $\dfrac{1}{5}$
D) None of these

Answer 1

Hint:The probability of an event can be found using total outcome and the favourable outcomes. We need to write all possible outcomes and identify our favourable outcomes. Then applying the formula we get the required probability.

Complete step-by-step answer:
Given that two unbiased coins are tossed. We want to calculate the probability of getting two heads. First, we can check what all cases happen when three coins are tossed.
We represent the output of the toss in an ordered triplet such that the first value represents the outcome of the first coin, and second and third likewise.
For convenience we write $H$ for getting head and $T$ for getting tail.
Then the possible cases are
$(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)$
In total there are $8$ cases.
Now we have to find our favourable cases, that is, those having at least two heads. Let the event be $A$.
At least two heads means either two heads or three heads.
So those cases are $(HHH),(HHT),(HTH),(THH)$.
There are $4$ cases.
So the probability of getting at least two heads is the number of outcomes with at least two heads divided by the total number of outcomes.
$\operatorname{P} (A) = \dfrac{4}{8} = \dfrac{1}{2}$

So, the correct answer is “Option A”.

Note:If the probability of an event happening is $p$, then the probability of not happening that event is $1 - p$.Probability of getting exactly two heads and at least two heads are different. Here there are only three events with exactly two heads. Then that probability becomes $\dfrac{3}{8}$.