Question

A dice is thrown twice. The probability of getting $4,5{\text{ or 6}}$ in the first throw and $1,2,3{\text{ or 4}}$ in the second throw is ?
A. $1$
B. $\dfrac{1}{3}$
C. $\dfrac{7}{{36}}$
D. None of these

Answer 1

Hint: In order to find the probability of the given statement, we would solve each condition separately that is for the first throw and for the second throw, finding their probability. And, since they are independent events, we would use the independent event formula of probability, substituting the values, solving and getting the results.

Formula used:
$P\left( S \right) = \dfrac{{n\left( S \right)}}{{n\left( E \right)}}$
\[P\left( {A \cap B} \right) = P\left( A \right).P\left( B \right)\]

Complete step by step answer:
We are given two throws, that is the dice is thrown twice.
From the Formulas of probability, we know that Probability is the ratio of number of favorable outcomes and the total number of favorable outcomes, that is numerically written as $P\left( S \right) = \dfrac{{n\left( S \right)}}{{n\left( E \right)}}$
Solving the two throws separately.

For the first throw:
Total number of outcomes $ = 6$ (As because a die has six numbers and any one can come at a time).
Possible outcomes $ = 3$ (As because we need about three numbers only that is $4,5{\text{ or 6}}$).
Substituting these in the Probability formula, we get:
$P\left( A \right) = \dfrac{3}{6}$ that is simplified and can be written as:
$\Rightarrow P\left( A \right) = \dfrac{1}{2}$

For the Second throw:
Total number of outcomes $ = 6$ (As because a die has six numbers and any one can come at a time). Possible outcomes $ = 4$ (Same reason, as because we need about four numbers only that can occur at a time that is $1,2,3{\text{ or 4}}$).
Substituting these in the Probability formula, we get:
$P\left( B \right) = \dfrac{4}{6}$ that is simplified and can be written as:
$\Rightarrow P\left( B \right) = \dfrac{2}{3}$

Since, the two throws are independent events that occur, so we use the Independent formula from probability that is \[P\left( {A \cap B} \right) = P\left( A \right).P\left( B \right)\].
Substituting the value of \[P\left( A \right){\text{ }}and{\text{ }}P\left( B \right)\] in the above formula, we get:
\[P\left( {A \cap B} \right) = \dfrac{1}{2} \times \dfrac{2}{3}\]
Cancelling out 2, we get:
\[P\left( {A \cap B} \right) = \dfrac{1}{3}\]
which is the required result. Therefore, the probability of getting $4,5{\text{ or 6}}$ in the first throw and $1,2,3{\text{ or 4}}$ in the second throw is \[\dfrac{1}{3}\].

Hence, option B is correct.

Note:Since, there were two separate throws of the dice, they are not thrown at once, and also shows that one’s value does not changed or affect the other’s value or in simple words they are not dependent on each other, that’s why they are called Independent events and we used this particular formula in our question.