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21 Oct 2023, 23:57
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A
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Difficulty:
45%
(medium)
Question Stats:
70% (02:20) correct
30%
(02:11)
wrong
based on 171
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In a game of dice, the participants need to roll a dice twice. If it is a prime number, then they receive $2. If it is a composite number, then they need to pay $3. If it is neither then they receive $4. What is the probability that a participant wins a net amount of $1?
A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{1}{3}\)
D. \(\frac{4}{9}\)
E. \(\frac{2}{3}\)
A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{1}{3}\)
D. \(\frac{4}{9}\)
E. \(\frac{2}{3}\)
22 Oct 2023, 02:24
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Snehaaaaa
To win a net amount of $1, the person should draw a composite number on one dice and 1 (neither prime nor composite) on another. The draw can be in any order.
Among {1, 2, 3, 4, 5, 6} we have two composite numbers.
Hence, number of possiblities = \(1 * 2 * 2 = 4\)
Probability = \(\frac{4}{36} = \frac{1}{9}\)
Option A
04 Feb 2025, 05:00
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
In a game of dice, the participants need to roll a dice twice. If it is a prime number, then they receive $2.
Prime numbers between 1 to 6 are 2, 3 and 5
=> We receive $2 if we get 2, 3 or 5
If it is a composite number, then they need to pay $3.
Composite numbers between 1 to 6 are 4 and 6
=> We need to pay $3 if we get 4 or 6
If it is neither then they receive $4.
Out of the numbers from 1 to 6, only number left is 1
=> We receive $4 if we get a 1
What is the probability that a participant wins a net amount of $1?
To get $1 we can have $4 - $3
=> Probability that we will get a 1 and we will get 4 or 6
=> P(1)*P(4,6) + P(4,6)*P(1) = \(\frac{1}{6} * \frac{2}{6}\) + \(\frac{2}{6} * \frac{1}{6}\) = \(\frac{2}{36}\) + \(\frac{2}{36}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
In a game of dice, the participants need to roll a dice twice. If it is a prime number, then they receive $2.
Prime numbers between 1 to 6 are 2, 3 and 5
=> We receive $2 if we get 2, 3 or 5
If it is a composite number, then they need to pay $3.
Composite numbers between 1 to 6 are 4 and 6
=> We need to pay $3 if we get 4 or 6
If it is neither then they receive $4.
Out of the numbers from 1 to 6, only number left is 1
=> We receive $4 if we get a 1
What is the probability that a participant wins a net amount of $1?
To get $1 we can have $4 - $3
=> Probability that we will get a 1 and we will get 4 or 6
=> P(1)*P(4,6) + P(4,6)*P(1) = \(\frac{1}{6} * \frac{2}{6}\) + \(\frac{2}{6} * \frac{1}{6}\) = \(\frac{2}{36}\) + \(\frac{2}{36}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
