Q24(i):
A die is thrown twice. What is the probability that (i) 5 will not come up either time? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution :

Given: A fair six-faced die is thrown twice. The possible outcomes for each throw are $\{1, 2, 3, 4, 5, 6\}$.

To Find: The probability that the number 5 will not come up in either of the two throws.

Step 1: Determining the Total Number of Possible Outcomes

When a die is thrown once, there are $6$ possible outcomes. When a die is thrown twice, the total number of outcomes is calculated by the product of the outcomes of each throw [Fundamental Counting Principle].

Total outcomes = $6 \times 6 = 36$.

The sample space $S$ consists of all ordered pairs $(a, b)$ where $a$ is the result of the first throw and $b$ is the result of the second throw:

$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),$

$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$

$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$

$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),$

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),$

$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

Step 2: Identifying Favorable Outcomes

Let $E$ be the event that 5 does not come up in either throw. This means that for any outcome $(a, b)$, $a \neq 5$ and $b \neq 5$.

If $a \neq 5$, then $a \in \{1, 2, 3, 4, 6\}$ (5 possibilities).

If $b \neq 5$, then $b \in \{1, 2, 3, 4, 6\}$ (5 possibilities).

The number of favorable outcomes $n(E)$ is the product of the number of choices for the first throw and the second throw:

$n(E) = 5 \times 5 = 25$.

Step 3: Calculating the Probability

The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes [Classical Definition of Probability].

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(E) = \frac{n(E)}{n(S)}$

$P(E) = \frac{25}{36}$

Alternative Method (Using Complementary Events):

Let $A$ be the event that 5 comes up at least once. The outcomes where 5 appears are:

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)$.

Counting these, we find $n(A) = 11$.

$P(A) = \frac{11}{36}$.

Since the event "5 does not come up" is the complement of event $A$ (denoted as $A'$), we use the property $P(A') = 1 - P(A)$:

$P(A') = 1 - \frac{11}{36} = \frac{36 - 11}{36} = \frac{25}{36}$.

Final Answer: The probability that 5 will not come up either time is $\frac{25}{36}$.

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Q24(i): A die is thrown twice. What is the probability that (i) 5 will not come up either time? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

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Q24(i):
A die is thrown twice. What is the probability that (i) 5 will not come up either time? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]