Question

In five throws with a single die what is the chance of throwing: (1) three aces exactly, (2) three aces at least?
A. $\dfrac{{123}}{{3888}}\,,\,\dfrac{{625}}{{648}}\,,$ Favourable way for event E when one number is selected are $(2,\;9)$
B. $\dfrac{{125}}{{3888}}\,,\,\dfrac{{23}}{{648}},$ Favourable way for event E when one number is selected are $(2,\;9)$
C. $\dfrac{{3763}}{{3888}}\,,\,\dfrac{{23}}{{648}},$ Favourable way for event E when one number is selected are $(2,\;9)$
D. $\dfrac{{3763}}{{3888}}\,,\,\dfrac{{625}}{{648}},$ Favourable way for event E when one number is selected are $(2,\;9)$

Answer 1

Here the given question needs to be solved using probability, in which we are going to use conditional probability formulae. Here we will see the different possibilities which are possible and then by using commutation formulae we can solve for the result needed in the question.

Formula used: $\Rightarrow P(X = n) = {}^y{C_n}{p^{n - z}}{q^z}$
Here $y$ is variable value assumed according to question and $p,q$ are conditional probabilities.

Complete step by step answer:
First, we have to assign a variable to the total number of rows. So, let us assume that the total number of rows is ‘n’. So:
$n = 5$
We will assume that the probability of getting an ace is ‘p’ and the probability for not getting an ice after throwing the dice once is ‘q’.
We know that the probability for getting an ace is:
$p = \dfrac{1}{6}$
Similarly, the probability for not getting an ace is:
$q = 1 - \dfrac{1}{6}$
$\Rightarrow q = \dfrac{5}{6}$
For finding the probability for getting the exact 3 aces has a formula. The formula is:
$P(X = 3) = {}^5{C_3}{p^3}{q^2}$
$\Rightarrow P(X = 3) = 10 \times {\left( {\dfrac{1}{6}} \right)^3} \times {\left( {\dfrac{5}{6}} \right)^2}$
$\Rightarrow P(X = 3) = \dfrac{{125}}{{3888}}$
For finding the probability for getting at least 3 aces is:
$\Rightarrow P(X = 3) + P(X = 4) + P(X = 5) = {}^5{C_3}{p^3}{q^2} + {}^5{C_4}{p^4}{q^1} + {}^5{C_5}{p^5}{q^0}$
$\Rightarrow P(X = 3) + P(X = 4) + P(X = 5) = 10 \times {\left( {\dfrac{1}{6}} \right)^3} \times {\left( {\dfrac{5}{6}} \right)^2} + 5 \times {\left( {\dfrac{1}{6}} \right)^4} \times \left( {\dfrac{5}{6}} \right) + {\left( {\dfrac{1}{6}} \right)^5}$
$\Rightarrow P(X = 3) + P(X = 4) + P(X = 5) = \dfrac{1}{{{6^5}}}(250 + 25 + 1)$
$\Rightarrow P(X = 3) + P(X = 4) + P(X = 5) = \dfrac{{23}}{{648}}$
So, our answer is $\dfrac{{125}}{{3888}}\,,\,\dfrac{{23}}{{648}},$ favourable way for event E when one number is selected are $(2,\;9)$

So, the correct answer is Option B.

Note: Probability means finding the possibility of any event out of the maximum possible events, here the given question needs to find the possibility of an event of coming aces in five roll of dice out of maximum possible events, and by using suitable formulae that is conditional probability we can find the result which is asked in the question.