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From a well shufflied pacl of 52 pl...

From a well shufflied pacl of 52 playing cards , four cards are accidently dropped . Find the probability that one card is missing from each suit .

A

`(17)/( 20825 )`

B

`(2197 )/( 20825 )`

C

`(197 )/( 1665)`

D

none of these

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The correct Answer is:
To solve the problem of finding the probability that one card is missing from each suit when four cards are dropped from a well-shuffled pack of 52 playing cards, we can follow these steps: ### Step 1: Understand the Composition of a Deck A standard deck of 52 playing cards consists of 4 suits: Hearts, Diamonds, Clubs, and Spades. Each suit contains 13 cards. ### Step 2: Define the Event We want to find the probability that exactly one card is missing from each of the four suits after dropping four cards. This means we need to select one card from each suit. ### Step 3: Calculate the Favorable Outcomes To find the number of favorable outcomes (i.e., the number of ways to choose one card from each suit), we can use the combination formula. For each suit, we can choose 1 card out of 13 available cards. Thus, the number of ways to choose one card from each suit is: \[ \text{Favorable Outcomes} = 13 \times 13 \times 13 \times 13 = 13^4 \] Calculating \( 13^4 \): \[ 13^4 = 28561 \] ### Step 4: Calculate the Total Outcomes Next, we need to determine the total number of ways to choose any 4 cards from the 52-card deck. This can be calculated using the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items (52 cards) and \( r \) is the number of items to choose (4 cards): \[ \text{Total Outcomes} = \binom{52}{4} = \frac{52!}{4!(52-4)!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} \] Calculating this gives: \[ \text{Total Outcomes} = \frac{52 \times 51 \times 50 \times 49}{24} = 270725 \] ### Step 5: Calculate the Probability Now, we can calculate the probability \( P \) that one card is missing from each suit: \[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{28561}{270725} \] ### Step 6: Simplify the Probability To simplify the fraction, we can check if both numbers have any common factors. In this case, they do not, so we can leave it as is. ### Final Answer Thus, the probability that one card is missing from each suit when four cards are dropped is: \[ P = \frac{28561}{270725} \]
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ARIHANT SSC-PROBABILITY-INTRODUCTORY EXERCISE -(20.1)
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  16. A bag contains 8 red and 4 green balls . Find the probabilty t...

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  17. A bag contains 8 red and 4 green balls . Find the probabilty t...

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  18. A bag contains 8 red and 4 green balls . Find the probabilty t...

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  20. The odds against of an event are 5:7 , find the probility of o...

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