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Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 23 maths textbook solution

Answers (1)

Answer:

X 0 1 2
P\left ( X \right ) \frac{19}{34} \frac{13}{34} \frac{1}{17}


Hint: use combination formula

Given: two cards are drawn simultaneously from a well shuffled deck of 52 cards. Find the probability distribution of the number of successes when getting a spade is considered a success

Solution: in a deck of 52 cards, there are 13 spades, let x be the random variable denoting the number of success and success here is getting a spade  for an event when two cards are drawn simultaneously

∴ We can say the number of successes is equal to some spades obtained in each draw.

∴ X can take value 0, 1 or 2

 [For selecting 0 spades, we removed all 13 spade from the deck and selected out of 39]

\therefore P(x=0)=\frac{39 c_{2}}{52 c_{2}}=\frac{39 \times 38}{52 \times 51}=\frac{19}{34} 

[For selecting 1 spades, we need to select and 1 out of 13 spade and not any other spade]

 P(x=1)=\frac{13 c_{1} \times 39 c_{1}}{52 c_{2}}=\frac{13 \times 39 \times 38}{52 \times 51}=\frac{13}{34} 

 [For selecting 2 spade, we need to select and 2 out of 13 spade]

P(x=2)=\frac{13 c_{2}}{52 c_{2}}=\frac{13 \times 12}{52 \times 51}=\frac{1}{17} 

Now we have  X and P(X)             

∴ Now we are ready to write the probability distribution of x

The following probability distribution is given in this table

X 0 1 2
P\left ( X \right ) \frac{19}{34} \frac{13}{34} \frac{1}{17}
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