Question

A coin is tossed 10 times. The probability of getting exactly six heads is
A. $\dfrac{512}{513}$
B. $\dfrac{105}{512}$
C. $\dfrac{100}{153}$
D. ${}^{10}{{C}_{6}}$

Answer 1

Hint: We first explain the concept of empirical probability and how the events are considered. We take the given events and find the number of outcomes. Using the probability theorem of $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$, we get the empirical probability of the shooting event.

Complete step-by-step solution:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome.
We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options.
We find the number of outcomes for both events. We take the conditional event A and the universal event as U and numbers will be denoted as $n\left( A \right)$ and $n\left( U \right)$.
We take the empirical probability of the given problem as $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$.
We are given that A coin is tossed 10 times, and we need the probability of finding 6 heads.
Therefore, if the condition of getting 6 heads is considered as A, then \[n\left( A \right)={}^{10}{{C}_{6}}=\dfrac{10!}{6!\times 4!}=210\].
The event U is to get results for 10 tosses. So, \[n\left( U \right)={{2}^{10}}\] as for every toss we have 2 options.
The empirical probability of the shooting event is $P\left( A \right)=\dfrac{210}{{{2}^{10}}}=\dfrac{105}{512}$.
The correct option is B.

Note: We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.