Question

The probability distribution of the number of doublets in four throws of a pair of dice is given by

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

To verify a probability distribution table quickly without computing every single cell, calculate the first entry ($X=0$) and the last entry ($X=4$). Here, $P(X=4) = (\frac{1}{6})^4 = \frac{1}{1296}$ and $P(X=0) = (\frac{5}{6})^4 = \frac{625}{1296}$. This lets you isolate the correct option immediately.

Answer 1

The Correct Option is D

Step 1: Define the experiment.
A pair of dice is thrown 4 times. Let $X$ count how many of these throws give a doublet. We want the full distribution of $X$.
Step 2: Find the chance of a doublet in one throw.
A pair of dice has $36$ outcomes; the doublets $(1,1)$ through $(6,6)$ are $6$ of them, so $p = \dfrac{6}{36} = \dfrac{1}{6}$ and $q = \dfrac{5}{6}$.
Step 3: Recognise the model.
The throws are independent with two outcomes each, so $X$ is binomial with $n = 4$: $P(X=r) = \binom{4}{r}\left(\tfrac{1}{6}\right)^r\left(\tfrac{5}{6}\right)^{4-r}$.
Step 4: Compute the low values.
$P(0) = \dfrac{625}{1296}$ and $P(1) = 4\cdot\dfrac{1}{6}\cdot\dfrac{125}{216} = \dfrac{500}{1296} = \dfrac{125}{324}$.
Step 5: Compute the middle value.
$P(2) = 6\cdot\dfrac{1}{36}\cdot\dfrac{25}{36} = \dfrac{150}{1296} = \dfrac{25}{216}$.
Step 6: Compute the high values and match.
$P(3) = \dfrac{20}{1296} = \dfrac{5}{324}$ and $P(4) = \dfrac{1}{1296}$. This complete list of probabilities is the table shown in option 4, agreeing with the key.
\[ \boxed{\text{Distribution in option 4}} \]