Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained.

Q. 17.     Suppose that two cards are drawn at random from a deck of cards. LetX be the number of aces obtained. Then the value of E(X) is Choose the correct answer in the following:

               (A)   \frac{37}{221}

               (B)    \frac{5}{13}

               (C)   \frac{1}{13}

                (D)  \frac{2}{13}

Answers (1)

best_answer

X be number od aces obtained.

X can be 0,1,2

There 52 cards and 4 aces, 48 are non-ace cards.

P(X=0)=P(0 \, ace\, and\, 2\, non\, ace\, cards)=\frac{^4C_2.^4^8C_2}{^5^2C_2}=\frac{1128}{1326}

P(X=1)=P(1 \, ace\, and\, 1\, non\, ace\, cards)=\frac{^4C_1.^4^8C_1}{^5^2C_2}=\frac{192}{1326}

P(X=2)=P(2 \, ace\, and\, 0\, non\, ace\, cards)=\frac{^4C_2.^4^8C_0}{^5^2C_2}=\frac{6}{1326}

The probability distribution is as :

X 0 1 2
P(X) \frac{1128}{1326} \frac{192}{1326} \frac{6}{1326}

 

E(X)=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}

E(X)=\frac{204}{1326}

E(X)=\frac{2}{13}

Option D is correct.

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question