A single coin is tossed $ 5 $ times. What is the probability of getting at least one head?
A. $ \dfrac{{27}}{{32}} $
B. $ \dfrac{3}{{32}} $
C. $ \dfrac{1}{{32}} $
D. $ \dfrac{{31}}{{32}} $

Hint: As we know that probability is the prediction of a particular outcome of a random event. It is a set of all the possible outcomes for a random experiment. We can calculate the question with the formula of probability i.e. Probability $ = \dfrac{{No.\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,\,outcomes}} $ . Let us assume a coin to be fair and $ 2 - $ sides. When we flip the coin five times, it has $ {2^5} = 32 $ outcomes. So we have the total number of outcomes $ = 32 $ .

Complete step-by-step answer:
We have the total number of cards $ = 32 $ .
We can show the outcomes using the Sample space then $ S = (H,H,H,H,H),(H,H,H,H,T),...,(T,H,H) $ .
So we can see that the probability of all tails in $ 5 $ throws is $ \dfrac{1}{{32}} $ .
We can say that the probability of getting at least one head $ = 1 - $ probability of one head.
So the required probability by putting value is $ 1 - \dfrac{1}{{32}} $ .
On solving it gives us $ \dfrac{{32 - 1}}{{32}} = \dfrac{{31}}{{32}} $
Hence the correct option is (d) $ \dfrac{{31}}{{32}} $ .
So, the correct answer is “Option D”.

Note: We should be careful that we have to find the probability of at- least one head , so we have to subtract it. If we see the complement of at least one tail means $ 0 $ tails or $ 5 $ heads, then we must surely see $ 1 $ outcome i.e. all heads. Then we can say that it must follow at least one tail that has $ 31 $ outcomes i.e. $ 32 - 1 = 31 $ .