Probability (1 Viewer)

[crex]

Sup guys, got a few questions:

1) The digits 1,2,3,4 are used to form numbers that may have 1,2,3 or 4 digits in them. If one of the numbers is selected at random, find the probability that: it is odd and greater than 3000 (Ans: 3/32)

2) A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining a pair (Ans: 352/833)

Thanks guise!

 

[HSC2014]

Where'd you get these questions? I keep getting a different ans for 1

 

[Sy123]

For 1). First find the number of numbers that we can make with only the digits (1, 2, 3, 4).

How many 1 digit numbers? 4
How many 2 digit numbers? 4*3 = 12
How many 3 digit numbers? 4*3*2 = 24
How many 4 digit numbers? 4*3*2*1 = 24

Total is 64.

Now find the number of numbers, that only have 1, 2, 3, 4 as their digits, and are greater than 3000. We can count them off of our fingers and get 6 easily, there is no real combinatorial way to get it, since there is a restriction, i.e. if we use 3 as the first digit, we cannot use 1 for the second since the number must be odd. We could divide the cases into whether 3 or 4 is the first digit, but that is too much of a hassle.

So by counting, there are 6 out of 64 cases, hence probability is 6/64=3/32

For 2.

What does it mean by 'pair'?

 

[braintic]

Sup guys, got a few questions:

1) The digits 1,2,3,4 are used to form numbers that may have 1,2,3 or 4 digits in them. If one of the numbers is selected at random, find the probability that: it is odd and greater than 3000 (Ans: 3/32)

2) A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining a pair (Ans: 352/833)

Thanks guise!

1. No. of 1-digit nos: 4
2. No. of 2-digit nos: 4×3=12
3. No. of 3-digit nos: 4×3×2=24
4. No. of 4-digit nos: 4×3×2×1=24
Total: 64

Must be a 4-digit no, first digit must be 3 or 4, last digit must be 1 or 3.
3 Cases: 3**1, 4**1, 4**3
Each case has 2 possibilities, so 6 possibilities in total.

So prob. = 6/64 = 3/32.

2. Total no. of hands = 52C5
Pick the denomination of the pair: 13 possibilities
Pick the 2 cards of that denomination from the 4 available: 4C2 possibilities
Pick the other 3 cards in order, making sure not to pick the same denomination twice, or the same denomination as the pair: 48×44×40, then divide by 3! to make order unimportant.

13×4C2×48×44×40÷3! / 52C5 = 352/833

 

[HSC2014]

ah, I was under the impression that for question one, the digits could be used multiple times t_t

 

[crex]

Right right
thanks sy123 and braintic