Question

Two coins are tossed together. What is the probability of getting exactly one head?

Answer 1

Let us denote S as the sample space, H as the head and T as the tail. Hence S=\left\\{ HH,HT,TH,TT \right\\} . From the sample space, we can see that the possible outcomes where one head appears when the two coins are tossed is \left\\{ HT,TH \right\\} . We will use the formula $\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ , where number of favourable outcomes $=2$ and total number of outcomes $=4$ .

Complete step-by-step answer:
We are given that two coins are tossed simultaneously. Let us find the sample space.
Let us denote S as the sample space, H as the head and T as the tail. Hence, the sample space is given by
S=\left\\{ HH,HT,TH,TT \right\\}
From the sample space, we can see that the possible outcomes where one head appears when the two coins are tossed is \left\\{ HT,TH \right\\} .
Hence, the number of favourable outcomes $=2$
Total number of outcomes $=4$
We know that $\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Let’s substitute the values in the above formula to get the probability of getting exactly one head.
$P\left( \text{exactly one head} \right)=\dfrac{\text{2}}{\text{4}}$
When we cancel the common factors from the numerator and denominator, we will get
$P\left( \text{exactly one head} \right)=\dfrac{\text{1}}{\text{2}}$
Hence, probability of getting exactly one head is $\dfrac{\text{1}}{\text{2}}$ .

Note: Always write the sample space to get the clear view of what to find. You may take \left\\{ HH \right\\} in the favourable outcomes which leads to wrong solutions. We will consider \left\\{ HH \right\\} when probability of at least one head is asked to find. We can see that the probability of getting exactly one head is the same as the probability of getting exactly one tail.