Q24(ii):
A die is thrown twice. What is the probability that (ii) 5 will come up at least once? [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 total number of possible outcomes for a single throw is $6$.

To Find: The probability that the number $5$ will come up at least once in the two throws.

Step 1: Determining the Total Number of Possible Outcomes

When a die is thrown twice, the total number of outcomes is calculated by the product of the outcomes of each throw. Since each throw has $6$ possible outcomes ($1, 2, 3, 4, 5, 6$):

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

The sample space $S$ is represented as follows:

(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 $5$ comes up at least once. This includes outcomes where $5$ appears on the first die, the second die, or both.

Outcomes where $5$ appears on the first die: $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$ [Total: $6$ outcomes].

Outcomes where $5$ appears on the second die: $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)$ [Total: $6$ outcomes].

Note: The outcome $(5,5)$ is common to both lists. To avoid double-counting, we list the unique favorable outcomes:

Favorable outcomes = $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)\}$.

Counting these, we find the number of favorable outcomes $n(E) = 11$.

Step 3: Calculating the Probability

The formula for the probability of an event $E$ is given by:

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

[Using the classical definition of probability]

$P(E) = \frac{11}{36}$

Step 4: Verification

Alternatively, we can use the complement rule. Let $E'$ be the event that $5$ never comes up. The outcomes that do not contain a $5$ are those where each die shows $1, 2, 3, 4,$ or $6$ ($5$ possibilities per die).

Number of outcomes without $5 = 5 \times 5 = 25$.

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

Since $P(E) = 1 - P(E')$, we have $P(E) = 1 - \frac{25}{36} = \frac{36-25}{36} = \frac{11}{36}$.

Final Answer: The probability that 5 will come up at least once is $\frac{11}{36}$.

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Q24(ii): A die is thrown twice. What is the probability that (ii) 5 will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

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Q24(ii):
A die is thrown twice. What is the probability that (ii) 5 will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]