Question

Which of the following is the probability of guessing correctly at least 7 out of 10 answers in a "True" or "False" test.
A. $\dfrac{11}{64}$
B. $\dfrac{11}{32}$
C. $\dfrac{11}{16}$
D. $\dfrac{27}{32}$

Answer 1

Hint: To find the probability of guessing correctly at least 7 out of 10 answers in a "True" or "False" test, we will find the probability of an answer being true, that is, \[P\left( T \right)=\dfrac{1}{2}\] and probability of an answer being false, that is, \[P\left( F \right)=\dfrac{1}{2}\] . Probability of guessing correctly at least 7 out of 10 answers in a "True" or "False" test can be written as $P\left( X\ge 7 \right)=P\left( X=7 \right)+P\left( X=8 \right)+P\left( X=9 \right)+P\left( X=10 \right)$ . We will use the formula $P\left( X=r \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$ to find the required probability.

Complete step by step answer:
We have to find the probability of guessing correctly at least 7 out of 10 answers in a "True" or "False" test.
We know that the probability of an event is given by
$P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Let us find the probability of an answer being true. We know that the total number of outcomes is 2 since either true or false can occur.
\[\Rightarrow P\left( T \right)=\dfrac{1}{2}\]
Now, let’s find the probability of an answer being false.
\[\Rightarrow P\left( F \right)=\dfrac{1}{2}\]
We have to find the probability of guessing correctly at least 7 out of 10 answers in a "True" or "False" test, that is, $P\left( X\ge 7 \right)$ .
We can write $P\left( X\ge 7 \right)$ as shown below.
$P\left( X\ge 7 \right)=P\left( X=7 \right)+P\left( X=8 \right)+P\left( X=9 \right)+P\left( X=10 \right)$
We know that the binomial distribution formula is for any random variable X, given by;
$P\left( X=r \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}...(i)$
where n is the number of experiments
r = 0, 1, 2, 3, 4, …
p = Probability of Success in a single experiment
q = Probability of Failure in a single experiment = 1 – p
Let us consider \[P\left( T \right)=p=\dfrac{1}{2}\] and $q=1-p=\dfrac{1}{2}=P\left( F \right)$
Let us now find $P\left( X\ge 7 \right)$ using (i).
$P\left( X\ge 7 \right)=P\left( X=7 \right)+P\left( X=8 \right)+P\left( X=9 \right)+P\left( X=10 \right)$
\[\Rightarrow P\left( X\ge 7 \right){{=}^{10}}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{7}}{{\left( \dfrac{1}{2} \right)}^{10-7}}{{+}^{10}}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{8}}{{\left( \dfrac{1}{2} \right)}^{10-8}}{{+}^{10}}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{9}}{{\left( \dfrac{1}{2} \right)}^{10-9}}{{+}^{10}}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}{{\left( \dfrac{1}{2} \right)}^{10-10}}\]
Let us now solve the powers. We will get
\[P\left( X\ge 7 \right){{=}^{10}}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{7}}{{\left( \dfrac{1}{2} \right)}^{3}}{{+}^{10}}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{8}}{{\left( \dfrac{1}{2} \right)}^{2}}{{+}^{10}}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{9}}{{\left( \dfrac{1}{2} \right)}^{1}}{{+}^{10}}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}{{\left( \dfrac{1}{2} \right)}^{0}}\]
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ and ${{a}^{0}}=1$ . Hence, the above equation can be written as
\[P\left( X\ge 7 \right){{=}^{10}}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{7+3}}{{+}^{10}}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{8+2}}{{+}^{10}}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{9+1}}{{+}^{10}}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}\]
Let us simplify the powers. We will get
\[P\left( X\ge 7 \right){{=}^{10}}{{C}_{7}}{{\left( \dfrac{1}{2} \right)}^{10}}{{+}^{10}}{{C}_{8}}{{\left( \dfrac{1}{2} \right)}^{10}}{{+}^{10}}{{C}_{9}}{{\left( \dfrac{1}{2} \right)}^{10}}{{+}^{10}}{{C}_{10}}{{\left( \dfrac{1}{2} \right)}^{10}}\]
From the above equation, we can see that \[{{\left( \dfrac{1}{2} \right)}^{10}}\] is common. Let us take it outside.
\[\Rightarrow P\left( X\ge 7 \right)=\left( ^{10}{{C}_{7}}{{+}^{10}}{{C}_{8}}{{+}^{10}}{{C}_{9}}{{+}^{10}}{{C}_{10}} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Let’s expand the above terms. We will get
\[P\left( X\ge 7 \right)=\left( \dfrac{10!}{7!\left( 10-7 \right)!}+\dfrac{10!}{8!\left( 10-8 \right)!}+\dfrac{10!}{9!\left( 10-9 \right)!}+\dfrac{10!}{10!\left( 10-10 \right)!} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
\[\Rightarrow P\left( X\ge 7 \right)=\left( \dfrac{10!}{7!3!}+\dfrac{10!}{8!2!}+\dfrac{10!}{9!1!}+\dfrac{10!}{10!0!} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
We know that 0!=1. Hence, the above equation becomes
\[P\left( X\ge 7 \right)=\left( \dfrac{10!}{7!3!}+\dfrac{10!}{8!2!}+\dfrac{10!}{9!1!}+\dfrac{10!}{10!} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
Let us expand the factorial. We will get
\[P\left( X\ge 7 \right)=\left( \dfrac{10!}{7!3!}+\dfrac{10!}{8\times 7!2!}+\dfrac{10!}{9\times 8\times 7!}+\dfrac{10!}{10\times 9\times 8\times 7!} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
We can take 10! Common from the numerator and 7! common from the denominator. We will get
\[P\left( X\ge 7 \right)=\left( \dfrac{1}{3!}+\dfrac{1}{8\times 2!}+\dfrac{1}{9\times 8}+\dfrac{1}{10\times 9\times 8} \right)\left( \dfrac{10!}{7!} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
Let us expand the factorials.
\[\Rightarrow P\left( X\ge 7 \right)=\left( \dfrac{1}{3\times 2\times 1}+\dfrac{1}{8\times 2\times 1}+\dfrac{1}{9\times 8}+\dfrac{1}{10\times 9\times 8} \right)\left( \dfrac{10\times 9\times 8\times 7!}{7!} \right){{\left( \dfrac{1}{2} \right)}^{10}}\]
Let’s simplify the denominators and cancel the common terms.
\[\Rightarrow P\left( X\ge 7 \right)=\left( \dfrac{1}{6}+\dfrac{1}{16}+\dfrac{1}{72}+\dfrac{1}{720} \right)10\times 9\times 8\times {{\left( \dfrac{1}{2} \right)}^{10}}\]
Let us take the LCM and simplify.
\[P\left( X\ge 7 \right)=\left( \dfrac{1\times 120}{6\times 120}+\dfrac{1\times 45}{16\times 45}+\dfrac{1\times 10}{72\times 10}+\dfrac{1}{720} \right)720\times {{\left( \dfrac{1}{2} \right)}^{10}}\]
Let us solve this.
\[\begin{align}
  & P\left( X\ge 7 \right)=\left( \dfrac{120+45+10+1}{720} \right)720{{\left( \dfrac{1}{2} \right)}^{10}} \\
 & \Rightarrow P\left( X\ge 7 \right)=\left( \dfrac{176}{720} \right)720{{\left( \dfrac{1}{2} \right)}^{10}} \\
\end{align}\]
Let us cancel the common factors. We will get
\[P\left( X\ge 7 \right)=176\times {{\left( \dfrac{1}{2} \right)}^{10}}\]
Let us expand \[{{\left( \dfrac{1}{2} \right)}^{10}}\] . We will get
\[\begin{align}
  & P\left( X\ge 7 \right)=\dfrac{176}{1024} \\
 & \Rightarrow P\left( X\ge 7 \right)=\dfrac{11}{64} \\
\end{align}\]

So, the correct answer is “Option A”.

Note: You may mistake when writing the formula for $P\left( X=r \right)$ as $^{n}{{C}_{r}}{{p}^{r}}{{q}^{n+r}}$ . Also, there can be mistake when writing the formula for $^{n}{{C}_{r}}$ as $\dfrac{n!}{r!\left( n+r \right)!}$ . You may write the probability of guessing correctly at least 7 out of 10 answers in a "True" or "False" test as $P\left( X\le 7 \right)$ which will lead to an incorrect solution. You must know the rules of exponents in order to solve the exponent numbers.