A pair of dice is rolled 5 times. let getting a total of 5 in a single throw is considered as success. If probability of getting atleast four success is \(\frac{x}{3}\) then x is equal to
A pair of dice is rolled 5 times. let getting a total of 5 in a single throw is considered as success. If probability of getting atleast four success is \(\frac{x}{3}\) then x is equal to
- \(\frac{41}{9^{5}}\)
- \(\frac{41}{9^{4}}\)
- \(\frac{123}{9^{5}}\)
- \(\frac{123}{9^{4}}\)
The Correct Option is C
Approach Solution - 1
P(atleast for the four success) = 5C4\((\frac{1}{9})^{4}\) . \(\frac{8}{9}\)+\((\frac{1}{9})^{3}\) = \(\frac{x}{3}\)
⇒ x = \(\frac{41\times 3}{9^{5}}\) = \(\frac{123}{9^{5}}\)
Approach Solution -2
Probability Calculation
Let S be the event of getting a sum of 5 when a pair of dice is thrown. The possible outcomes for getting a sum of 5 are {(1, 4), (2, 3), (3, 2), (4, 1)}.
So, the number of favorable outcomes is 4.
Total possible outcomes when two dice are thrown = 6 * 6 = 36
$P(\text{Success}) = P(\text{Getting a sum of 5}) = \frac{4}{36} = \frac{1}{9}$
$P(\text{Failure}) = 1 - P(\text{Success}) = 1 - \frac{1}{9} = \frac{8}{9}$
We are given that the dice are thrown 5 times, and we want to find the probability of at least 4 successes. This can be calculated as: $P(\text{At least 4 successes}) = P(\text{4 successes}) + P(\text{5 successes})$
Using the binomial probability formula: $P(X = r) = \binom{n}{r} p^r (1 - p)^{n-r}$, where n is the number of trials, r is the number of successes, and p is the probability of success.
$P(\text{4 successes}) = \binom{5}{4} (\frac{1}{9})^4 (\frac{8}{9})^1 = 5 \times \frac{1}{9^4} \times \frac{8}{9} = \frac{40}{9^5}$
$P(\text{5 successes}) = \binom{5}{5} (\frac{1}{9})^5 (\frac{8}{9})^0 = 1 \times \frac{1}{9^5} \times 1 = \frac{1}{9^5}$
$P(\text{At least 4 successes}) = \frac{40}{9^5} + \frac{1}{9^5} = \frac{41}{9^5} = \frac{41}{59049}$
We are given that the probability of at least 4 successes is $\frac{k}{3^{11}}$. Since $9^5 = (3^2)^5 = 3^{10}$, we have $\frac{41}{9^5} = \frac{41}{3^{10}} = \frac{41 \times 3}{3^{11}} = \frac{123}{3^{11}} $.
Given that the probability of at least 4 successes is $\frac{k}{3^{11}}$, we have:
$\frac{k}{3^{11}} = \frac{123}{3^{11}}$
Conclusion: Therefore, $k = 123$.
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Concepts Used:
Probability
Probability is defined as the extent to which an event is likely to happen. It is measured by the ratio of the favorable outcome to the total number of possible outcomes.
The definitions of some important terms related to probability are given below:
Sample space
The set of possible results or outcomes in a trial is referred to as the sample space. For instance, when we flip a coin, the possible outcomes are heads or tails. On the other hand, when we roll a single die, the possible outcomes are 1, 2, 3, 4, 5, 6.
Sample point
In a sample space, a sample point is one of the possible results. For instance, when using a deck of cards, as an outcome, a sample point would be the ace of spades or the queen of hearts.
Experiment
When the results of a series of actions are always uncertain, this is referred to as a trial or an experiment. For Instance, choosing a card from a deck, tossing a coin, or rolling a die, the results are uncertain.
Event
An event is a single outcome that happens as a result of a trial or experiment. For instance, getting a three on a die or an eight of clubs when selecting a card from a deck are happenings of certain events.
Outcome
A possible outcome of a trial or experiment is referred to as a result of an outcome. For instance, tossing a coin could result in heads or tails. Here the possible outcomes are heads or tails. While the possible outcomes of dice thrown are 1, 2, 3, 4, 5, or 6.