Question

A six-faced dice is so biased that it is twice as likely to show an even number as an odd number when thrown. It is thrown twice. The probability that the sum of two numbers thrown is even, is $A)\dfrac{1}{{12}}$
$B)\dfrac{1}{6}$
$C)\dfrac{1}{3}$
$D)\dfrac{5}{9}$

Answer 1

Hint: First, we need to know the concept of probability.
Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of outcomes.
Since a dice has six faces and it is thrown twice.
If the sum of the two numbers needs to be even, then we need to multiply two even numbers or two odd numbers.
Suppose if we multiply an odd number with the even number then it gets the odd number but which is not the requirement in the given question.
Formula used:
$P = \dfrac{F}{T}$where P is the probability, F is the possible favorable events and T is the total outcomes from the given.
Suppose $A$ is an odd number then $2A$is an even number (the number $2$multiplies with any odd number and the resultant gets the even number like $2 \times 3 = 6$)

Complete step by step answer:
Let us assume the probability of getting the odd numbers to be $A$ and from this, we can say that the probability of getting the even number is $2A$(where $2$into any odd numbers gets the even numbers only)
Hence, in the probability concept the total sum of the given events that will not exceed $1$ (this is the most popular concept that used in the probability that the total fraction will not exceed $1$and everything will be calculated under the number $0 - 1$as zero is the least possible outcome and one is the highest outcome)
Thus, the sum of the probability of the odd number and even numbers probability is $A + 2A = 1$
Further solving we get, $3A = 1 \Rightarrow A = \dfrac{1}{3}$
Thus, the probability of getting the odd numbers is written in the fraction as $\dfrac{1}{3}$
Apply the value of odd numbers in the equation that even number is $2A$, then we get $2A = 2(\dfrac{1}{3}) \Rightarrow \dfrac{2}{3}$ and thus $\dfrac{2}{3}$is the even number probability.
From the given question as they say, the sum of the two numbers thrown that needs to be an even number and the possibility to get that is either both numbers are odd or both numbers are even.
Starting with both odd numbers, that is $\dfrac{1}{3}$thrown twice and we get $\dfrac{1}{3} \times \dfrac{1}{3} \Rightarrow \dfrac{1}{9}$is the probability that dice thrown twice and getting the outcome as even numbers (from both odd numbers)
Similarly, with both even numbers, that is $\dfrac{2}{3}$thrown twice and we get $\dfrac{2}{3} \times \dfrac{2}{3} \Rightarrow \dfrac{4}{9}$is the probability that dice thrown twice and getting the outcome as even numbers (from both even numbers)
Therefore, the total probability is $\dfrac{1}{9} + \dfrac{4}{9} = \dfrac{5}{9}$

So, the correct answer is “Option D”.

Note: Since if we add the total number of getting the even numbers and odd numbers then we get $\dfrac{1}{3} + \dfrac{2}{3} = 1$which is the concept used to solve the problem that the sum of the fractions needs to equal to $1$
If we divide the probability value and multiply with the $100$then we get the actual percentage of the given value.
Like take the odd number probability that is $\dfrac{1}{3}$and by the use of division operation we get $\dfrac{1}{3} = 0.33...$and multiply with the number $100$and thus we get $0.33 \times 100 = 33.33\% $which is the percentage of getting the odd number.