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An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.
Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018
- a)0.84
- b)0.69
- c)0.16
- d)0.52
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An event A is considered to be rolling a fair dice .If 10 dice are rol...
Let Xi denote the ith dice.
Since events are independent,
E(xi) = 7/2
V(xi) = 350/12
According to Central Limit Theorem,
P(30≤ X ≤40) = P(30 – 35 ≤ X – 35 ≤ 40-35)
Its converting to standard form, X – E[X] / std, where std is standard deviation.
Thus,
P = 2N(1.0184) – 1
= 0.69
Most Upvoted Answer
An event A is considered to be rolling a fair dice .If 10 dice are rol...
To find the probability that the sum obtained on the faces of 10 rolled dice is between 30 and 40 (inclusive), we can use the concept of generating functions.
Let's define the generating function for a single dice as:
\( f(x) = x + x^2 + x^3 + x^4 + x^5 + x^6 \)
The generating function for 10 dice will be the product of the generating functions for each dice:
\( F(x) = f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \)
We are interested in finding the probability of the sum being between 30 and 40. Let's define a new generating function \( G(x) \) that represents the sum of the faces on the dice:
\( G(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)^{10} \)
The coefficient of \( x^k \) in \( G(x) \) will give us the number of ways to obtain a sum of \( k \) on the 10 dice. To find the probability, we need to calculate the sum of the coefficients of \( x^k \) for \( k \) from 30 to 40 (inclusive), and divide it by the total number of possible outcomes.
Using the binomial theorem, we can expand \( G(x) \) as:
\( G(x) = \binom{10}{0}x^{0} + \binom{10}{1}x^{1} + \binom{10}{2}x^{2} + ... + \binom{10}{10}x^{10} \)
To find the desired probability, we need to calculate the sum of the coefficients of \( x^{30} \) to \( x^{40} \).
Calculating the sum of the coefficients can be a tedious task, but we can make use of the fact that the sum of the coefficients of a generating function is equal to the value of the generating function when \( x = 1 \).
So, to find the desired probability, we need to evaluate \( G(1) \), which will give us the sum of the coefficients.
Substituting \( x = 1 \) in \( G(x) \):
\( G(1) = \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + ... + \binom{10}{10} \)
Using the binomial coefficient formula, we can simplify the expression:
\( G(1) = 2^{10} \)
Now, the total number of possible outcomes is \( 6^{10} \) since each dice has 6 faces.
Therefore, the probability is given by:
\( \frac{G(1)}{6^{10}} = \frac{2^{10}}{6^{10}} \)
Simplifying the expression:
\( \frac{2^{10}}{6^{10}} = \left(\frac{2}{6}\right)^{10} = \left(\frac{1}{3}\right)^{10} \)
Calculating \( \left(\frac
Let's define the generating function for a single dice as:
\( f(x) = x + x^2 + x^3 + x^4 + x^5 + x^6 \)
The generating function for 10 dice will be the product of the generating functions for each dice:
\( F(x) = f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \cdot f(x) \)
We are interested in finding the probability of the sum being between 30 and 40. Let's define a new generating function \( G(x) \) that represents the sum of the faces on the dice:
\( G(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)^{10} \)
The coefficient of \( x^k \) in \( G(x) \) will give us the number of ways to obtain a sum of \( k \) on the 10 dice. To find the probability, we need to calculate the sum of the coefficients of \( x^k \) for \( k \) from 30 to 40 (inclusive), and divide it by the total number of possible outcomes.
Using the binomial theorem, we can expand \( G(x) \) as:
\( G(x) = \binom{10}{0}x^{0} + \binom{10}{1}x^{1} + \binom{10}{2}x^{2} + ... + \binom{10}{10}x^{10} \)
To find the desired probability, we need to calculate the sum of the coefficients of \( x^{30} \) to \( x^{40} \).
Calculating the sum of the coefficients can be a tedious task, but we can make use of the fact that the sum of the coefficients of a generating function is equal to the value of the generating function when \( x = 1 \).
So, to find the desired probability, we need to evaluate \( G(1) \), which will give us the sum of the coefficients.
Substituting \( x = 1 \) in \( G(x) \):
\( G(1) = \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + ... + \binom{10}{10} \)
Using the binomial coefficient formula, we can simplify the expression:
\( G(1) = 2^{10} \)
Now, the total number of possible outcomes is \( 6^{10} \) since each dice has 6 faces.
Therefore, the probability is given by:
\( \frac{G(1)}{6^{10}} = \frac{2^{10}}{6^{10}} \)
Simplifying the expression:
\( \frac{2^{10}}{6^{10}} = \left(\frac{2}{6}\right)^{10} = \left(\frac{1}{3}\right)^{10} \)
Calculating \( \left(\frac
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An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer?
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An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer?.
An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An event A is considered to be rolling a fair dice .If 10 dice are rolled each of which are independent, find the probability that the sum obtained on the faces is between 30 and 40 both inclusive.Given: N(1.0184) = 0.84 , which is value of normal distribution for X = 1.018a)0.84b)0.69c)0.16d)0.52Correct answer is option 'B'. Can you explain this answer?.
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