Question

An experiment succeeds twice as often as it fails. Find the probability that in 4 trails there will be at least three success
A. $\dfrac{4}{{27}}$
B. $\dfrac{8}{{27}}$
C. $\dfrac{{16}}{{27}}$
D. $\dfrac{{24}}{{27}}$

Answer 1

Here the given question is based on the concept of probability. We have to find the probability of at least 3 successes in 4 trails. For this first we have to find the probability of success and failure. Further by using a binomial distribution i.e., $P\left( {X = x} \right){ = ^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$, to get the required probability.

Complete step by step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted
with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to
happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible
one and 1 indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the
number of favourable outcomes and the total number of outcomes.
$\text{Probability of event to happen}\,P\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total Number of outcomes}}}$
Consider the question:
Given,
The probability of success is twice the probability of failure
The number of trails $n = 4$
Let us take ‘$p$’ to be the probability of success and ‘$q$’ be the probability of failure.
Probability of success $p = 2q$ ----(1)
Probability of failure $q = 1 - p$ ----(2)
On substituting ‘$q$’ in equation (1) we have
$\Rightarrow \,\,\,p = 2\left( {1 - p} \right)$
$\Rightarrow \,\,\,p = 2 - 2p$
Add $2p$ on both the sides, we have
$\Rightarrow \,\,\,p + 2p = 2$
$\Rightarrow \,\,\,3p = 2$
Divide both side by 3, then we get
$\Rightarrow \,\,\,p = \dfrac{2}{3}$
$\therefore$ Probability of success $p = \dfrac{2}{3}$
Substitute the $p$ value in equation (2), then
$\Rightarrow \,\,\,q = 1 - \dfrac{2}{3}$
Take 3 as LCM in RHS
$\Rightarrow \,\,\,q = \dfrac{{3 - 2}}{3}$
$\Rightarrow \,\,\,q = \dfrac{1}{3}$
$\therefore$ Probability of failure $q = \dfrac{1}{3}$
Let $X$ be the random variable that represents the number of successes in four trials. It’s a Bernoulli trial so, $X$ has a binomial distribution, we obtain
$P\left( {X = x} \right){ = ^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$ ----(3)
Probability of at least 3 success $= P\left( {X \geqslant 3} \right)$
On putting values of $p$, $q$, $n$ and $x$ in equation (3), then
$P\left( {X = x} \right){ = ^4}{C_3}{\left( {\dfrac{2}{3}} \right)^3}{\left( {\dfrac{1}{3}} \right)^{4 - 3}} + {\,^4}{C_4}{\left( {\dfrac{2}{3}} \right)^4}{\left( {\dfrac{1}{3}} \right)^{4 - 4}}$
$\Rightarrow \,\,\,P\left( {X = x} \right){ = ^4}{C_3}{\left( {\dfrac{2}{3}} \right)^3}{\left( {\dfrac{1}{3}} \right)^1} + {\,^4}{C_4}{\left( {\dfrac{2}{3}} \right)^4}{\left( {\dfrac{1}{3}} \right)^0}$
We know the formula of combination $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, then
$\Rightarrow \,\,\,P\left( {X = x} \right) = \dfrac{{4!}}{{(4 - 3)!3!}}\left( {\dfrac{{{2^3}}}{{{3^3}}}} \right)\left( {\dfrac{1}{3}} \right) + \,\dfrac{{4!}}{{(4 - 4)!4!}}\left( {\dfrac{{{2^4}}}{{{3^4}}}} \right)\left( 1 \right)$
$\Rightarrow \,\,\,P\left( {X = x} \right) = \dfrac{{4!}}{{(1)!3!}}\left( {\dfrac{{{2^3}}}{{{3^4}}}} \right) + \,\dfrac{{4!}}{{(0)!4!}}\left( {\dfrac{{{2^4}}}{{{3^4}}}} \right)$
$\Rightarrow \,\,\,P\left( {X = x} \right) = \dfrac{{4 \times 3!}}{{(1)3!}}\left( {\dfrac{{{2^3}}}{{{3^4}}}} \right) + \,\dfrac{{4!}}{{(1)4!}}\left( {\dfrac{{{2^4}}}{{{3^4}}}} \right)$
$\Rightarrow \,\,\,P\left( {X = x} \right) = 4\left( {\dfrac{{{2^3}}}{{{3^4}}}} \right) + \,\left( {\dfrac{{{2^4}}}{{{3^4}}}} \right)$
$\Rightarrow \,\,\,P\left( {X = x} \right) = {2^2}\left( {\dfrac{{{2^3}}}{{{3^4}}}} \right) + \,\left( {\dfrac{{{2^4}}}{{{3^4}}}} \right)$
We know the property of exponent ${a^m}{a^n} = {a^{mn}}$, then
$\Rightarrow \,\,\,P\left( {X = x} \right) = \left( {\dfrac{{{2^5}}}{{{3^4}}}} \right) + \,\left( {\dfrac{{{2^4}}}{{{3^4}}}} \right)$
Convert the exponent number to the whole number, then we have
$\Rightarrow \,\,\,P\left( {X = x} \right) = \left( {\dfrac{{32}}{{81}}} \right) + \,\left( {\dfrac{{16}}{{81}}} \right)$
On simplification, we get
$\Rightarrow \,\,\,P\left( {X = x} \right) = \dfrac{{32 + 16}}{{81}}$
$\Rightarrow \,\,\,P\left( {X = x} \right) = \dfrac{{48}}{{81}}$
Divide both numerator and denominator by 3, we get
$\therefore \,\,\,P\left( {X = x} \right) = \dfrac{{16}}{{27}}$
Hence the required probability is $\dfrac{{16}}{{27}}$

So, the correct answer is “Option C”.

Note: The probability is a number of possible values. Candidate must know we have to use
permutation concept or combination concept to solve the given problem because it is the first and
main thing to solve the problem.
$1 - P\left( A \right)$ refers to the probability that Event A will not occur is denoted by $P\left( {A'} \right)$.
Remember, factorial is the continued product of first n natural numbers is called the “n factorial “ and it represented by $n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot 3 \cdot 2 \cdot 1$.