Question

A die is thrown three times. Getting a 3 or 6 is considered a success. What is the probability of at least two successes?
A. \[\dfrac{2}{9}\]
B. \[\dfrac{7}{{27}}\]
C. \[\dfrac{1}{{27}}\]
D. None of these

Answer 1

Hint: First we will find the probability of getting a 3 or 6. The probability of getting 3 or 6 is the probability of success. Then by using the formula \[P\left( {E'} \right) = 1 - P\left( E \right)\] we find the probability of failure. Since the die is thrown three times. So we find the probability of getting success 2 times and 3 times by using the formula binomial distribution.

Formula used
\[P\left( {E'} \right) = 1 - P\left( E \right)\]
Binomial distribution:
\[p\left( {r} \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}\]
Where \[p\] is the probability of success, \[q\] is the probability of failure.

Complete step by step solution
Given that a die is thrown three times and getting a 3 or 6 is considered a success.
A die has 6 faces.
The total number of possible outcomes is 6.
The number of favourable outcomes is 2.
The probability of success is \[\dfrac{2}{6} = \dfrac{1}{3}\].
Apply the formula \[P\left( {E'} \right) = 1 - P\left( E \right)\] to get the probability of failure
The probability of failure is \[1 - \dfrac{1}{3} = \dfrac{2}{3}\]
Since at least 2 successes are required and the die is thrown three times.
There is a possibility that all three times get success.
Apply the formula of binomial distribution to calculate the probability for two successes out of 3:
Now \[n = 3\], \[r = 2\], \[p = \dfrac{1}{3}\], and \[q = \dfrac{2}{3}\]
\[P\left( {2} \right) = {}^3{C_2}{\left( {\dfrac{1}{3}} \right)^2}\left( {\dfrac{2}{3}} \right)\]
 Apply the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
\[ = \dfrac{{3!}}{{1!2!}}{\left( {\dfrac{1}{3}} \right)^2}\left( {\dfrac{2}{3}} \right)\]
\[ = \dfrac{{3 \times 2!}}{{1!2!}}{\left( {\dfrac{1}{3}} \right)^2}\left( {\dfrac{2}{3}} \right)\]
\[ = 3 \times \dfrac{2}{{{3^3}}}\]
\[ = \dfrac{2}{9}\]
Apply the formula of binomial distribution to calculate the probability for three successes out of 3:
Now \[n = 3\], \[r = 3\], \[p = \dfrac{1}{3}\], and \[q = \dfrac{2}{3}\]
\[P\left( {3} \right) = {}^3{C_3}{\left( {\dfrac{1}{3}} \right)^3}{\left( {\dfrac{2}{3}} \right)^0}\]
 Apply the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
\[ = \dfrac{{3!}}{{1!3!}}{\left( {\dfrac{1}{3}} \right)^3}{\left( {\dfrac{2}{3}} \right)^0}\]
\[ = 1 \times {\left( {\dfrac{1}{3}} \right)^3} \times 1\]
\[ = \dfrac{1}{{27}}\]
Now we will add both probabilities to getting the required probability
Th required probability is \[\dfrac{2}{9} + \dfrac{1}{{27}}\]
\[ = \dfrac{{6 + 1}}{{27}}\]
\[ = \dfrac{7}{{27}}\]
Hence option B is the correct option.

Note: Many students often do a mistake to apply the binomial distribution. They applied \[b\left( {n,p} \right) = {}^n{C_r}{\left( p \right)^{n - r}}{\left( q \right)^r}\] which is an incorrect formula. The correct formula of binomial distribution is \[b\left( {n,p} \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}\].