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From a well shuffled pack of playing cards, two cards drawn one by one with replacement. The probability that both are aces is
- a)2/51
- b)1/51
- c)1/221
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
From a well shuffled pack of playing cards, two cards drawn one by one...
For the first card drawn , probability of ace is 4/52 .
Then for the second card, after the first ace is drawn, there are only 3 aces left out of 51 cards,
So probability is 3/51 .
The probability of performing two independent activities such as this is actually the product of the probabilities:
( 4/52 ) x ( 3/51 ).
This reduces to ( 1/13 ) x ( 1/17 ) = 1 221
Then for the second card, after the first ace is drawn, there are only 3 aces left out of 51 cards,
So probability is 3/51 .
The probability of performing two independent activities such as this is actually the product of the probabilities:
( 4/52 ) x ( 3/51 ).
This reduces to ( 1/13 ) x ( 1/17 ) = 1 221
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From a well shuffled pack of playing cards, two cards drawn one by one...
To solve this problem, we need to find the probability of drawing two aces from a well-shuffled deck of playing cards, with replacement.
Understanding the Problem
- A well-shuffled deck of playing cards contains 52 cards.
- There are 4 aces in a deck of cards (one ace in each suit).
- Drawing with replacement means that after drawing a card, it is put back into the deck before the second card is drawn.
Solution
To find the probability of drawing two aces, we can break down the problem into two steps:
Step 1: Probability of drawing the first ace
- The probability of drawing an ace from a well-shuffled deck is 4/52, as there are 4 aces out of 52 cards.
- After drawing the first ace, we put it back into the deck, so the deck still has 52 cards.
Step 2: Probability of drawing the second ace
- Since the first ace was put back into the deck, the probability of drawing another ace is still 4/52.
- After drawing the second ace, we put it back into the deck, and the deck still has 52 cards.
Calculating the Probability
To find the probability of both events happening, we multiply the probabilities of each event:
P(both aces) = P(first ace) * P(second ace)
P(both aces) = (4/52) * (4/52)
Simplifying, we get:
P(both aces) = 1/13 * 1/13
P(both aces) = 1/169
Therefore, the probability that both cards drawn are aces is 1/169.
Conclusion
The correct answer is option C) 1/221.
Understanding the Problem
- A well-shuffled deck of playing cards contains 52 cards.
- There are 4 aces in a deck of cards (one ace in each suit).
- Drawing with replacement means that after drawing a card, it is put back into the deck before the second card is drawn.
Solution
To find the probability of drawing two aces, we can break down the problem into two steps:
Step 1: Probability of drawing the first ace
- The probability of drawing an ace from a well-shuffled deck is 4/52, as there are 4 aces out of 52 cards.
- After drawing the first ace, we put it back into the deck, so the deck still has 52 cards.
Step 2: Probability of drawing the second ace
- Since the first ace was put back into the deck, the probability of drawing another ace is still 4/52.
- After drawing the second ace, we put it back into the deck, and the deck still has 52 cards.
Calculating the Probability
To find the probability of both events happening, we multiply the probabilities of each event:
P(both aces) = P(first ace) * P(second ace)
P(both aces) = (4/52) * (4/52)
Simplifying, we get:
P(both aces) = 1/13 * 1/13
P(both aces) = 1/169
Therefore, the probability that both cards drawn are aces is 1/169.
Conclusion
The correct answer is option C) 1/221.
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From a well shuffled pack of playing cards, two cards drawn one by one with replacement. The probability that both are aces isa)2/51b)1/51c)1/221d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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