Question

An Unbiased cubic die marked with \[1,2,2,3,3,3,\] is rolled 3 times. The probability of getting a total score of 4 or 6 is
A. \[\dfrac{16}{216}\]
B. \[\dfrac{50}{216}\]
C. \[\dfrac{60}{216}\]
D. None of these

Answer 1

Hint: In general probability of an event (P(E)) can be calculated by formula as given below: $P(E)=\dfrac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}$
Also we need to first make possible outcomes for getting sum 4 or 6 and then we can calculate probability for particular cases.

Complete step by step solution:
The numbers marked on the die are \[1,2,2,3,3,3,\]
So the probability of getting the number 1 is \[\dfrac{1}{6}\] since there is only I on the dice.
The probability of getting the number 2 on the dice is \[\dfrac{2}{6}\]since there are two 2s on the dice.
The probability of getting the number 3 on the dice is \[\dfrac{3}{6}\] since there are three 3s on the dice.
The chances of getting the sum 4 when the die I is rolled three times: \[(\text{1 1 2})\]
! now this combination can be arranged in
\[\dfrac{3!}{2!}=3\]ways
We write\[2!\] in the denominator because two numbers are repeated.
The chances of getting 6 when die is rolled 3 times is:
(i). \[(1,2,3)\]
Now this combination can be arranged in \[3!\]ways i.e. 6 ways
(ii). \[(2,2,2)\]
Now this combination can be arranged only in 1 way because all are identical numbers that means \[\dfrac{3!}{3!}=1\]way.
Now the probability of getting sum 4 is
\[\Rightarrow \dfrac{1}{6}\times \dfrac{1}{6}\times \dfrac{2}{6}\times (3)\] ……. 1
The probability of getting sum 6 is
\[\Rightarrow \dfrac{1}{6}\times \dfrac{2}{6}\times \dfrac{3}{6}\times (6)+\dfrac{2}{6}\times \dfrac{2}{6}\times \dfrac{2}{6}\times (1)\] …… 2
The numbers in the bracket denotes the number of ways it can be arranged.
So the total probability is:
Adding 1 and 2
\[\Rightarrow \dfrac{1}{6}\times \dfrac{1}{6}\times \dfrac{2}{6}\times (3)+\dfrac{1}{6}\times \dfrac{2}{6}\times \dfrac{3}{6}\times (6)+\dfrac{2}{6}\times \dfrac{2}{6}\times \dfrac{2}{6}\times (1)\]
\[\Rightarrow \dfrac{50}{216}\]

Note: Along with writing probability we should also mention the number of ways the combination can be arranged and according to that we can calculate the probability of that particular event.