Question

If one coin is tossed 10 times, find the probability of: -
1. exactly 6 heads
2. at least 6 heads
3. at most 6 heads

Answer 1

Hint: Assume ‘X’ as the number of heads appearing. Apply Bernoulli probability distribution formula given by: - \[P\left( X=a \right)={}^{n}{{C}_{a}}{{p}^{a}}{{q}^{n-a}}\] to calculate the probability of getting ‘a’ number of heads. Here, ‘n’ is the number of times the coin tossed, ‘a’ is the number of heads appearing, ‘p’ is the probability of getting ‘a’ head and ‘q’ is the probability of not getting ‘a’ head. For part 1 substitute X = a = 6, for part 2 substitute \[X=a\ge 6\] and for part 3 substitute \[X=a\le 6\].

Complete step-by-step solution
Here, we have been given that a coin is tossed 10 times and we have to find the probability of getting exactly six heads, at least six heads, and at most six heads.
Now, the above question is related to Bernoulli’s probability distribution. So, applying the formula of Bernoulli probability distribution for getting X = a heads, we have,
\[\Rightarrow P\left( X=a \right)={}^{n}{{C}_{a}}{{p}^{a}}{{q}^{n-a}}\]
Here, p = probability of getting a head.
q = probability of not getting a head.
n = number of times the coin is tossed.
Now, when a coin is tossed one time we may get either a head or a tail. So favorable outcome for getting a head is 1 and the total outcome is 2. Therefore,
\[\Rightarrow p=\dfrac{1}{2},q=1-\dfrac{1}{2}=\dfrac{1}{2}\] and n = 10
Now, the required probabilities can be calculated as follows: -
1. The probability of getting exactly 6 heads.
\[\begin{align}
  & \Rightarrow P\left( X=6 \right) \\
 & \Rightarrow {}^{10}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{6}}\times {{\left( \dfrac{1}{2} \right)}^{10-6}} \\
\end{align}\]
\[\Rightarrow {}^{10}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}\]
\[\Rightarrow 210\times {{\left( \dfrac{1}{2} \right)}^{10}}\]
2. The probability of getting at least 6 heads
\[\begin{align}
  & \Rightarrow P\left( X\ge 6 \right) \\
 & \Rightarrow P\left( X=6 \right)+P\left( X=7 \right)+P\left( X=8 \right)+P\left( X=9 \right)+P\left( X=10 \right) \\
 & \Rightarrow {}^{10}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{6}}{{\left( \dfrac{1}{2} \right)}^{10-6}}+{}^{10}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{7}}{{\left( \dfrac{1}{2} \right)}^{10-7}}+{}^{10}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{8}}{{\left( \dfrac{1}{2} \right)}^{10-8}}+{}^{10}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{9}}{{\left( \dfrac{1}{2} \right)}^{10-9}}+{}^{10}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}{{\left( \dfrac{1}{2} \right)}^{10-10}} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {}^{10}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}+{}^{10}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}} \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{10}}\left[ {}^{10}{{C}_{6}}+{}^{10}{{C}_{7}}+{}^{10}{{C}_{8}}+{}^{10}{{C}_{9}}+{}^{10}{{C}_{10}} \right] \\
 & \Rightarrow 386{{\left( \dfrac{1}{2} \right)}^{10}} \\
\end{align}\]
3. The probability of getting at most 6 heads.
\[\begin{align}
  & \Rightarrow P\left( X\le 6 \right) \\
 & \Rightarrow P\left( X=1 \right)+P\left( X=2 \right)+P\left( X=3 \right)+P\left( X=4 \right)+P\left( X=5 \right)+P\left( X=6 \right) \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{10}}\left[ {}^{10}{{C}_{1}}+{}^{10}{{C}_{2}}+{}^{10}{{C}_{3}}+{}^{10}{{C}_{4}}+{}^{10}{{C}_{5}}+{}^{10}{{C}_{6}} \right] \\
 & \Rightarrow 847{{\left( \dfrac{1}{2} \right)}^{10}} \\
\end{align}\]

Note: One may note that it will be very difficult to solve the above problem without using Bernoulli probability distribution rules. We have used p and q for denoting the probability of success and failure respectively because they are used as general notation in the formula. One can use different notations but do not get confused in the formula. ‘p’ must be raised to the power ‘a’ and ‘q’ to the power (n – a).