Question

8 coins are tossed simultaneously. The probability of getting at least 6 heads are:
(a) $\dfrac{57}{64}$
(b) $\dfrac{229}{256}$
(c) $\dfrac{7}{64}$
(d) $\dfrac{37}{256}$

Answer 1

Hint: We know that the formula for the probability is equal to the ratio of favorable outcomes to the total outcomes. Total outcomes in this problem is calculated as follows: 8 coins are tossed simultaneously so for each coin two possibilities are there so for 8 coins, total possibilities are ${{2}^{8}}$. And the favorable outcomes contain at least 6 heads, this means that 6 heads, 7 heads and 8 heads should be there in the toss so we have to find the ways in which when 8 coins are tossed then 6 heads, 7 heads or 8 heads can come.

Complete step by step solution:
In the above problem, it is given that 8 coins are tossed simultaneously and we have to find the probability when at least 6 heads are there.
The formula for the probability is equal to the ratio of favorable outcomes to the total outcomes.
$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Now, we need the value of favorable and total outcomes.
Total outcomes is calculated as follows:
It is given that 8 coins tossed simultaneously so for each coin two possibilities are there then for 8 coins the total possibilities are equal to ${{2}^{8}}$. Writing this expression 2 to the power 8 in the compact form we get,
$\begin{align}
  & 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\
 & =256 \\
\end{align}$
Hence, the total number of outcomes is equal to 256.
Now, we are going to find favorable outcomes. For that, we have to find the number of possibilities when at least 6 heads are there so we need to find the number of ways in which out of 8 coins, 6 coins have heads, 7 coins have heads and 8 coins have heads.
The number of possibilities when 6 heads are present amongst 8 coins:
HHHHHHTT
The above arrangement of coins is one of the ways so basically we have to find the arrangement of the above heads and tails which is equal to:
$\dfrac{8!}{6!2!}$
Now, we can write $8!$ as $8.7.6!$ in the above expression and we get,
$\dfrac{8.7.6!}{6!2!}$
We can write $2!$ as 2.1 and $6!$ will be cancelled out from the numerator and the denominator and we get,
$\begin{align}
  & \dfrac{8.7}{2.1} \\
 & =\dfrac{56}{2} \\
 & =28 \\
\end{align}$
The number of possibilities when 7 heads and 2 tails are there:
$\begin{align}
  & \dfrac{8!}{7!1!} \\
 & =\dfrac{8.7!}{7!} \\
\end{align}$
In the above expression, $7!$ will be cancelled out from the numerator and the denominator and we get,
8
The number of possibilities when all the 8 coins have heads then only 1 possibility is possible which is equal to:
HHHHHHHH
Now, adding all the three cases we get,
$\begin{align}
  & 28+8+1 \\
 & =37 \\
\end{align}$
Hence, the favorable outcomes are equal to 37.
Now, substituting the value of total outcomes and favorable outcomes in the probability formula we get,
$\dfrac{37}{256}$

So, the correct answer is “Option d”.

Note: The mistake that could be possible in the above problem is that while finding the number of ways when 6 heads or 7 heads will be there then you might forget to arrange the possibilities in each of the two cases.
For e.g. when 6 heads will occur then one of the possible arrangement is equal to:
HHHHHHTT
And you just consider this one case. So, make sure you will consider all the arrangements of the 6 heads and 2 tails.