Question

Find the probability distribution of the number of doublets in three throws of a pair of dice.

Answer 1

Hint: In this question understand the meaning of doublets which means 2 same numbers at the same time on both the dice, find the total possible doublets possible use this information to get the better approach toward the solution to the problem.

Complete step-by-step answer:
According to the question, two dice are thrown, so the outcomes we will get are 6 $\times$ 6 = 36 outcomes. We are given that number of doublets which means we get the same numbers on both throws of the dice. So, the number of doublets possible while 2 throws of dice are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
Now on this account we will have two situations:
Probability of getting a doublet
Probability of not getting a doublet
Here, probability of getting a doublet is= number of total doublets/total possible outcome = $\dfrac{6}{{36}}$ = $\dfrac{1}{6}$
Now, the second possibility, probability of not getting a doublet is = $1 - \dfrac{1}{6} = \dfrac{5}{6}$
According to the question we have thrown two dice thrice which means we can acquire either 0 doublet or 1 doublet or 2 doublet or 3 doublet.
Let us then consider the value of doublet be X.
From the above conclusion value of X can be = 0, 1, 2, 3
Let’s first find out the probability of getting 0 doublets, which can be written as:
As we have already concluded the probability of not getting a doublet which is =$\dfrac{5}{6}$
And the two dice were thrown 3 times, Hence P (X = 0) = P (0 doublet on three thrown)
$\Rightarrow$ P (0 doublet)$\times$ P (0 doublet)$\times$ P (0 doublet)
Substituting the values in the above formula
$\Rightarrow$ $\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6}$
$\Rightarrow$ $\dfrac{{125}}{{216}}$
Now, the probability when X = 1
P (X = 1) = P (one doublet on three throw)
As the two dice is thrown three times, the probability we will get each time will be added
Which will then = P( 1 doublet) $\times$ P ( 0 doublet ) $\times$ P(0 doublet)+ P(0 doublet) $\times$ P(1 doublet)$\times$ P(0 doublet)+P(0 doublet) $\times$ P(0 doublet) $\times$ P(1 doublet)
Substituting the values in the above formula
$\dfrac{1}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} + \dfrac{5}{6} \times \dfrac{1}{6} \times \dfrac{5}{6} + \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$
$\Rightarrow$ $3 \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$
$\Rightarrow$ $\dfrac{{75}}{{216}}$
Similarly for P(X = 2)
P (X = 2) = P (Two doublet on three throws)
$\Rightarrow$P(one doublet) $\times$ P(one doublet) $\times$ P(0 doublet) + P(one Doublet) $\times$ P(0 doublet) $\times$ P(one doublet)+P(0 doublet)$\times$ P(one doublet) $\times$ (one doublet)
Substituting the values in the above formula
$\dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{5}{6} + \dfrac{1}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{1}{6} \times \dfrac{1}{6}$
$\Rightarrow$ $3 \times \dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{5}{6}$
$\Rightarrow$ $\dfrac{{15}}{{216}}$
Now, lastly we have to find out the probability (X = 3)
For three doublets each doublet should be one doublet
$\Rightarrow$P (three doublets on three throws)
$\Rightarrow$P (One doublet) $\times$ P (one doublet) $\times$ P (one doublet) = $\dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{1}{6}$ = $\dfrac{1}{{216}}$
So, the probability distribution of the number of doublets is $\dfrac{1}{{216}}$

Note: In the above question the concept of probability can be explained as the method to find the chances of outcome of any event so let’s take an example to explain this concept suppose you have 1 dice and you want a 6 in one throw so the chances of occurring of this event is calculated by the method of probability, so in this case we know that the total number outcomes are 6 and in 6 outcomes there is only one chance to get 6 so the chance of getting 6 is given by $\dfrac{1}{6}$ .