Question

# A coin is tossed 40 times and it shows its tail 24 times. To find the probability of getting a head. \begin{align} & \text{a) }\dfrac{2}{5} \\ & \text{b) }\dfrac{3}{5} \\ & \text{c) }\dfrac{1}{2} \\ & \text{d) }\dfrac{17}{40} \\ \end{align}

Hint: For this question we will first note that there are only two outcomes for a coin tossed. It is either heads or tails. To find the required probability we will use the formula of probability which is $\dfrac{\text{number of required outcomes}}{\text{Total outcomes}}$ .

Now it is given that a coin is tossed 40 times and among the 40 times it showed tail 24 times
Now the result of a coin toss can either be head or tail hence if it showed tail 24 times means that the remaining outcomes were heads.
This means out of 40 outcomes, If 24 outcomes were tail then 40 – 24 outcomes were head.
Hence, 16 outcomes were head. ……(1)
Now let B be the event such that we get ahead. Then from equation (1) we get
$n(B)=16.............(2)$
Now we know that the formula of probability is $\dfrac{\text{number of required outcomes}}{\text{Total outcomes}}$
Now from equation (2) we know the number of required outcomes is 16.
Also the total outcome is 40
Hence applying these formula we get $p(B)=\dfrac{16}{40}$
Let us reduce the fraction by dividing the numerator and denominator by 8.
Hence, $p(B)=\dfrac{2}{5}$
Hence we get the probability of getting a head is $\dfrac{2}{5}$
So, the correct answer is “Option A”.

Note: While taking the ratio of probability note that we need to find heads and hence find the number of outcomes in which the coin showed heads. Since the outcome of tail is given, do not directly substitute the value in Formula. Instead find the required outcome by subtracting the number of times it showed tails from the total number of times the coin was tossed.