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Sindya draws two cards at random from a pack of cards. After returning them to the pack and shuffling it Bindya draws two cards at random. The probability that they have exactly one common cards is
- a)25/663
- b)28/663
- c)50/663
- d)2/663
- e)None of the Above
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Sindya draws two cards at random from a pack of cards. After returnin...
Sindya can draw any two cards. Now Bindya has to choose exactly one same card of Sindya's card. This can be done in 2C1 ways.
Remaining 1 card can be drawn in 50C1 ways
So, number of favorable cases = 2 × 50
Total number of case = 50C2 = 26×51
Most Upvoted Answer
Sindya draws two cards at random from a pack of cards. After returnin...
To find the probability that Sindya and Bindya have exactly one common card, we need to consider two scenarios:
1. Sindya draws a card, and Bindya does not draw the same card.
2. Sindya does not draw a card, and Bindya draws the same card.
Let's calculate the probability for each scenario:
Scenario 1:
1. The probability that Sindya draws a card is 1 (since there are 52 cards in the pack and she can choose any card).
2. The probability that Bindya does not draw the same card is (51/52) (since there are 51 cards remaining in the pack and only one of them is the same card Sindya drew).
Therefore, the probability for scenario 1 is (1 * 51/52).
Scenario 2:
1. The probability that Sindya does not draw a card is (51/52) (since there are 51 cards remaining in the pack and she did not draw any).
2. The probability that Bindya draws the same card is (1/52) (since there is only one card in the pack that is the same as the one Sindya did not draw).
Therefore, the probability for scenario 2 is ((51/52) * (1/52)).
Since the two scenarios are mutually exclusive (they cannot happen at the same time), we can add their probabilities to find the total probability that Sindya and Bindya have exactly one common card:
Total probability = (1 * 51/52) + ((51/52) * (1/52))
= 51/52 + 51/2704
= (51 * 2704 + 52 * 51) / 52 * 2704
= (137904 + 2652) / 140608
= 140556 / 140608
= 50/663
Therefore, the probability that Sindya and Bindya have exactly one common card is 50/663, which corresponds to option C.
1. Sindya draws a card, and Bindya does not draw the same card.
2. Sindya does not draw a card, and Bindya draws the same card.
Let's calculate the probability for each scenario:
Scenario 1:
1. The probability that Sindya draws a card is 1 (since there are 52 cards in the pack and she can choose any card).
2. The probability that Bindya does not draw the same card is (51/52) (since there are 51 cards remaining in the pack and only one of them is the same card Sindya drew).
Therefore, the probability for scenario 1 is (1 * 51/52).
Scenario 2:
1. The probability that Sindya does not draw a card is (51/52) (since there are 51 cards remaining in the pack and she did not draw any).
2. The probability that Bindya draws the same card is (1/52) (since there is only one card in the pack that is the same as the one Sindya did not draw).
Therefore, the probability for scenario 2 is ((51/52) * (1/52)).
Since the two scenarios are mutually exclusive (they cannot happen at the same time), we can add their probabilities to find the total probability that Sindya and Bindya have exactly one common card:
Total probability = (1 * 51/52) + ((51/52) * (1/52))
= 51/52 + 51/2704
= (51 * 2704 + 52 * 51) / 52 * 2704
= (137904 + 2652) / 140608
= 140556 / 140608
= 50/663
Therefore, the probability that Sindya and Bindya have exactly one common card is 50/663, which corresponds to option C.
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Sindya draws two cards at random from a pack of cards. After returning them to the pack and shuffling it Bindya draws two cards at random. The probability that they have exactly one common cards isa)25/663b)28/663c)50/663d)2/663e)None of the AboveCorrect answer is option 'C'. Can you explain this answer?
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