Question

Jersey-Cash-5 is a New Jersey Lottery that \benets education and institutions". In this lottery, a
game is played by picking 5 distinct numbers out of numbers 1 to 40.
Compute the number of ways and the probability of winning each of the following cases. Show the
combinatorial derivation. Use a calculator to perform the nal step in the computation. Express the
nal result as a ratio 1=x where the denomintor x is rounded to the nearest integer.
(a) Jackpot: Match 5 out of 5.
(b) Second Prize: Match 4 out of 5. (You win $500.)
(c) Third Prize: Match 3 out of 5. (You win $11.)

Answer 1

Please ans this question. exam in couple hours n this is likey to be on the exam !!

Answer 2

@amistre64 and @TuringTest any idea abt this question ?

Answer 3

what is the prob of picking any one number correctly?

Answer 4

@amistre64 how do u define correctly ? i dint understand ur question

Answer 5

if this is a fair game, there are 40 distinct elements to choose at random; so each element must have the same probability of being picked ... any given element therefore has a 1/40 probability of being chosen

right?

Answer 6

mm yes

Answer 7

there is a success rate "s" of picking a correct number (1/40)
this means that there is also a failure rate "f" of (39/40)

there is a setup of choosing "n" numbers from these odds as:

\[(s+f)^n\]

Answer 8

this has been worked out as equivalent to saying:\[\Large\binom{n}{k}s^{k}f^{n-k}\]so in this case
\[\Large\binom{5}{k}~\frac{1}{40}^{k}~\frac{39}{40}^{5-k}\]
should give you the probability of choosing n numbers correctly out of 5

Answer 9

let k be the number of correct choices that is

Answer 10

i am not familiar with those formulas. can u tell me what are they and when are they used ?

Answer 11

they are used for a discrete random variable that only has 2 likely outcomes - its called a binomial

Answer 12

i cant see a simpler way to present it without having to write up a chapters worth of text :/

Answer 13

hehe i understand ! is there any other way to do this problem without using this formulas?

Answer 14

there are a certain number of ways to pick 5 elements; each draw is either a success or a failure

5C0 or 5C5
fffff sssss

5C1 or 5C4
ffffs ssssf
fffsf sssfs
ffsff ssfss
fsfff sfsss
sffff fssss

5C2 or 5C3
fffss sssff
ffssf ssffs
fssff sffss
ssfff ffsss
ffsfs ssfsf
fsfsf sfsfs
sfsff fsfss
fsffs sfssf
sffsf fssfs
sfffs fsssf

32 actual outcomes

Answer 15

the probability of getting any one of these is the product of their s and f values:

s = 1/40
f = 39/10

this pretty much sums up the formula

Answer 16

39/40 that is

Answer 17

notice that there are 5C0 ways of getting none correct, s^0 and f^5

(5 0) (1/40)^0 (39/40)^5


notice that there are 5C5 ways of getting all correct, s^5 and f^0

(5 5) (1/40)^5 (39/40)^0


notice that there are 5C2 ways of getting 2 correct, s^2 and f^5-2

(5 2) (1/40)^2 (39/40)^3

etc ....

Answer 18

that makes more sense now !! thanks a lot !

Answer 19

good luck :)

Answer 20

@amistre64 sry 1 more question

consider a deck of 53 cards, which includes a joker (wild card). The possible poker hands are:
(a) Five-of-a-kind (Four of one kind, plus a joker)
(b) Straight Flush (ve cards same suit and in consecutive order).
(c) Four-of-a-Kind (4,1)
(d) Full House (3,2)
(e) Flush
(f) Straight
(g) Three-of-a-Kind (3,1,1)
(h) Two Pairs (2,2,1)
(i) One Pair (2,1,1,1)

Answer 21

(j) High Card: This occurs when none of the above hands is made.
For example, a Four-of-a-Kind hand can be made in two ways:
Without joker: four of one kind plus one of another kind.
With a joker: three of one kind, plus a joder, plus one of another kind.
Compute the number of ways to get each hand, and the probability.
Check your work by adding the number of ways for all cases. The total must be C(53;5).