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8. Two cards are drawn at random from an ordinary deck of 52 cards. Find the probability that all cards are diamonds if no replacement is done. A. 1/13 B. 3/52 C. 1/5 D. 1/17
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Gauthmathier8037
Grade 10 · 2022-03-24
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8. Two cards are drawn at random from an ordinary deck of
8. Two cards are drawn at random from an ordinary - Gauthmath cards. Find the probability that all cards are diamonds if no replacement is done.
A.
\frac {1}{13}
B.
\frac {3}{52}
C.
\frac {1}{5}
D.
\frac {1}{17}
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A card is drawn at random from an ordinary deck of 52 playing cards. The probability that the card is a 10 or a spade is:
- \(\rm \dfrac{17}{52}\)
- \(\rm \dfrac{7}{39}\)
- \(\rm \dfrac{4}{13}\)
- \(\rm \dfrac{3}{13}\)
Answer (Detailed Solution Below)
Option 3 : \(\rm \dfrac{4}{13}\)
Free
Electric charges and coulomb's law (Basic)
10 Questions 10 Marks 10 Mins
Concept:
- The probability of occurrence of an event A out a total possible outcomes N, is given by P(A) = \(\rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which event A can occur.
- For two events A and B we have P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
Calculation:
Let's say that A is the event of drawing a 10 and B be the event of drawing a spade.
Probability of drawing a 10 = P(A) = \(\rm \dfrac{4}{52}\).
Probability of drawing a spade = P(B) = \(\rm \dfrac{13}{52}\).
Probability of drawing a 10 of spade = P(A ∩ B) = \(\rm \dfrac{1}{52}\).
∴ Probability of drawing a 10 or a spade = P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= \(\rm \dfrac{4}{52}+\dfrac{13}{52}- \dfrac{1}{52}=\dfrac{16}{52}=\dfrac{4}{13}\).
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Video Transcript
Here the first 1 is possibility of 2 cod hens, so which is 2 c 2, which is 42 factorial, divided by 2 factorial multiplied by 51 factorial, which is 1326, then second, 1, getting 2 years is the possibility of getting 2 s, which is equal to 2 Out of 4, so 4 c 2 divided by 2 c 2, which is 0.0045, then second, 1, all spades. So totally we have 13 spades out of this. We are selecting 2 s, 13 c 2, divided by 52 c 2, which is 13 factorial by 2. Factorial multiplied by 11 factorial divided by 52 factorial by 2, factorial multiplied by 50 factorial, which is 0.0508, then third, 1 onelphase cord. So totally we have 12 phase cots. Therefore, the probabilities 12 c 2 divided by 52 c 2, which is 0.0 49 double 7 point. Then fourth 1 is at least 1 yes, so we can have us at least 1, yes, for 1 multiplied. By from the remaining, we can have 1 cot divided by 52 c 2, plus both the 4 c 2 divided by 52 c. 2. Point upon simplification. We may get it as 0.14932 and the last 1 is 2 cots of same soot, so either it may be a spade or diamond of horten or something so we can or clever. So we can write us, which is 4 times 13 c 2, divided by 52 c 2. Upon simplification, we may get the value for this as 0.23529.