Probability Help? (1 Viewer)

[jadenmaccas]

I got stuck in a question from the 2007 past Mathematics paper from the BOS.

(b)A pack of 52 cards consists of four suits with 13 cards in each suit.

(i) One card is drawn from the pack and kept on the table. A second card is drawn and placed beside it on the table. What is the probability that the second card is from a different suit to the first?
(ii) The two cards are replaced and the pack shuffled. Four cards are chosen from the pack and placed side by side on the table. What is the probability that these four cards are all from different suits?

Anyone can explain how to do it showing clear working?

 

[BadMeetsEvil]

1) P of same suit for second draw = 12/51
Therefore P of different suit= 1-12/51 = ... Work it out
2) P different suit = (13/52 x 13/51 x 13/50 x 13/49)

 

[Aesytic]

not entirely sure if this is right

(i) the probability of getting a card of one suit initially is 1, and since the card isn't replaced, the probability of the second card being a different suit is (13*3)/51 = 13/17 (three different suits to choose from each with 13 cards divided by 1 less card than the initial number)

.'. the probability of both cards being different suits is 1 * 13/17 = 13/17

(ii) the probability of the first card being of a certain suit is 1, the probability of the second card being a different suit is 13/17 from (i), the probability of the third card being another suit is (13*2)/50 = 13/25 (since there are 26 cards with a suit that hasn't been chosen yet), and the probability of the fourth card being a different suit is 13/49 (only 1 suit that hasn't been chosen yet).

.'. the probability of all 4 cards being different suits is 1 * 13/17 * 13/25 * 13/49 = 2197/20825

the value seems weird though, so i think i might be wrong

 

[Sy123]

I got stuck in a question from the 2007 past Mathematics paper from the BOS.

(b)A pack of 52 cards consists of four suits with 13 cards in each suit.

(i) One card is drawn from the pack and kept on the table. A second card is drawn and placed beside it on the table. What is the probability that the second card is from a different suit to the first?
(ii) The two cards are replaced and the pack shuffled. Four cards are chosen from the pack and placed side by side on the table. What is the probability that these four cards are all from different suits?

Anyone can explain how to do it showing clear working?

b)
i)The probability of the first card getting a certain suit is 1/4
The probability of having the second card the same suit is 12/51

ii)
Name the first suit the 'n' suit

The probability of the first card being the n suit is 1/4 (13/52)
The probability of the second card being the 'y' suit is 13/51
The probability of the third card being the 'x' suit is 13/50
The probability of the forth card being the 'z' suit is 13/49

1/4 x 13/51 x 13/50 x 13/49 = 0.0043857 or 0.4% or 2197/49980

This really doesnt look right, and by the way Im using the Windows calculator so yeah that is why I have no simplified fraction