Solution to Question 01:

3 cards can be drawn from a pack of 52 cards in 52C3 ways:
= (52 * 51 * 50) / (3 *2 *1) ways

= 26 * 17 * 50 ways

Let A = event of getting an ace card, a jack card and a queen card
There are 4 ace cards, 4 jack cards and 4 queen cards

1 ace card can be drawn in 4C1 ways = 4 ways
1 jack card can be drawn in 4C1 ways = 4 ways
1 queen card can be drawn in 4C1 ways = 4 ways

Therefore, n(A) = 4 * 4 * 4 ways

Probability of this event = (4*4*4)/(26*17*50) = 16/5525

Question - 2: 4 cards are drawn at random from a deck of cards. find the probability of getting all the four cards of different numbers...


Total number of way - 52C4
Favourable:
There are 10 number cards of each suite = 40

first card can be selected in 40C1 ways
second card can be selected in 36C1 ways
third - 32C1
forth - 28C1

4 cards can be selected in 40 * 36 * 32* 28 ways

But this has taken the order of selection into account
e.g.,

1 2 3 4
2 1 3 4

have been taken as 2 options in the above counting and thus divide by 4!

thus fovourable = (40 * 36 * 32* 28)/4!

divide favourable by total to get the answer

thanks

kartikshah wrote:Solution to Question 01:

3 cards can be drawn from a pack of 52 cards in 52C3 ways:
= (52 * 51 * 50) / (3 *2 *1) ways

= 26 * 17 * 50 ways

Let A = event of getting an ace card, a jack card and a queen card
There are 4 ace cards, 4 jack cards and 4 queen cards

1 ace card can be drawn in 4C1 ways = 4 ways
1 jack card can be drawn in 4C1 ways = 4 ways
1 queen card can be drawn in 4C1 ways = 4 ways

Therefore, n(A) = 4 * 4 * 4 ways ace j, a

Probability of this event = (4*4*4)/(26*17*50) = 16/5525

if for selecting an ace, a jack and a king i do,
12c1*8c1*4c1 what answer will this give to me?

abhishek07sep wrote:
if for selecting an ace, a jack and a king i do,
12c1*8c1*4c1 what answer will this give to me?

If you go this route, you can still get the right answer, as long as you do the division the correct way.
Here are both the questions solved using this:

1st question:
Required probability = (12C1*8C1*4C1)/(52C1*51C1*50C1) = (12*8*4)/(52*51*50) = 16/5525

2nd question: In the same way as above,
Required probability = (40*36*32*28)/(52*51*50*49)

Let me know if this helps