Question

An unbiased die is tossed twice. Find the probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss:
(a). \[\dfrac{1}{3}\]
(b). \[\dfrac{2}{3}\]
(c). \[\dfrac{1}{2}\]
(d). \[\dfrac{1}{4}\]

Answer 1

Hint: We will use the formula to determine probability of favorable outcomes. The probability of a favorable event is given by the number of favorable outcomes dividing by total number of possible outcomes.

Complete step-by-step solution -

Given an unbiased die is tossed twice. We have to find the probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss.
When a dice is tossed, total possible outcomes is given by,
 {1,2,3,4,5,6}
Then the number of total possible outcomes is given by 6.
When the dice is tossed twice the total Possible Outcome is given by,
6² = 36.
Therefore, we get the total number of possible outcomes in our case is 36.
The favourable Outcome = A = {4,5,6} and B = {1,2,3,4}.
Probability of an event \[=\dfrac{\text{total favorable outcomes}}{\text{total possible outcomes}}\]
\[\begin{align}
  & P(A)=\dfrac{3}{6} \\
 & \Rightarrow P(A)=\dfrac{1}{2} \\
\end{align}\]
And Probability for B is given by,
\[\begin{align}
  & P(B)=\dfrac{4}{6} \\
 & \Rightarrow P(B)=\dfrac{2}{3} \\
\end{align}\]

Probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss is given by multiplying P(A) with P(B).
Then the required probability is given by,
\[P(A) \times P(B)=\left( \dfrac{2}{3} \right)\left( \dfrac{1}{2} \right)\]
\[\Rightarrow P(A) \times P(B)=\dfrac{1}{3}\]
Therefore, we get the required probability as \[\dfrac{1}{3}\].
Matching from the option given above we have option (a) is correct.

Note: The possibility of error in this question can be at the point where you are calculating both the probability of the even A and the event B separately. Always go for calculating separately and multiplying them if the probability is dependent on each other.