Probability Question
 

Find the probability for the following event:

A king, ace, jack of clubs or 10 of diamonds appears in drawing a single card from a well-shuffled ordinary deck of 52 cards.

[QUOTE=aimtoachieve;725776]Find the probability for the following event:

A king, ace, jack of clubs or 10 of diamonds appears in drawing a single card from a well-shuffled ordinary deck of 52 cards.[/QUOTE]

There are 4 Kings, 4 Aces, 1 Jack of clubs and 1 10 of Diamonds.

4+4+1+1=10

P(A king, ace, jack of clubs or 10 of diamonds)= 10/52 = 5/26

[QUOTE=Buddha;725787]There are 4 Kings, 4 Aces, 1 Jack of clubs and 1 10 of Diamonds.

4+4+1+1=10

P(A king, ace, jack of clubs or 10 of diamonds)= 10/52 = 5/26[/QUOTE]

O... now, I get it. I thought the questioner was talking about the king, ace and jack of clubs and 1 10 of diamonds. By the way, in that case, the answer would be 4/52, right?

[QUOTE=aimtoachieve;725906]O... now, I get it. I thought the questioner was talking about the king, ace and jack of clubs and 1 10 of diamonds. By the way, in that case, the answer would be 4/52, right?[/QUOTE]

Yes, in that case it would be 1/13

Statistics Paper 2014
 

[B]Statistics Paper 2014, Question No. 2[/B]

(i) 8!
(ii) Every couple is now a single entity. There are 4 total entities. Hence, answer = 4!
(iii) All men are now a single entity. Total 5 entities, hence 5! ways. However 4 men can also be made to sit in 4! ways among themselves. Hence, answer = 5!*4! .

Correction:

(ii) Every couple is now a single entity. There are 4 total entities. However, within a couple, man and woman may sit in different order. What would be the answer? 4!*(2^4)?

[QUOTE=aimtoachieve;726176]Correction:

(ii) Every couple is now a single entity. There are 4 total entities. However, within a couple, man and woman may sit in different order. What would be the answer? 4!*(2^4)?[/QUOTE]

Correct

A man invited 5 friends. He was born in April as also all the invited friends. What is the probability that none of the friends was born on the same day of the month as the host?

Lets say the host was born on 1st April. The probability of this is 1/30. Now, none of the friends should have their D.O.B. as 1st April, hence probability of D.O.B. of any one friend is 29/30 as he can be born on any of the remaining 29 days. However, there are 5 friends, hence, (29/30)^5. However, the host can be born on any of the 30 days. Hence, final answer: [(1/30)*(29/30)^5]*30.

Correct?

[QUOTE=aimtoachieve;726486]A man invited 5 friends. He was born in April as also all the invited friends. What is the probability that none of the friends was born on the same day of the month as the host?

Lets say the host was born on 1st April. The probability of this is 1/30. Now, none of the friends should have their D.O.B. as 1st April, hence probability of D.O.B. of any one friend is 29/30 as he can be born on any of the remaining 29 days. However, there are 5 friends, hence, (29/30)^5. However, the host can be born on any of the 30 days. Hence, final answer: [(1/30)*(29/30)^5]*30.

Correct?[/QUOTE]

The probability of a friend sharing his birthday with host = 1/30

The probability of five friends sharing their birthday with host, P(H) = (1/30)^5

Probability that none were born on the same day of the month as host = 1 - P(H)

can u refer good book for probability