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Probability Questions

chee

New member
Joined
May 3, 2012
Messages
12
Need help with probability questions i failed on my test.............

Q1: the game of two handed cribbage is played with a standard 52 card deck. Each player is dealt hands of 6 cards.

[a]. how many different 6 card hands could be dealt in euchre?

for this question i did 52 factorial divided by 46 factorial but got the answer wrong [pure guess]
is it 52! divided by (46!)(6!) ?

. In order to have the possibility of having the highest hand in cribbage you must be dealt three 5s and one Jack and two other cards. The other cards cannot contain a 5(but could be a Jack). How many possible 6 card hands could be dealt which would give you the possibility of getting a highest hand?

this question i have no clue..

[c]. What is the probability of being deal a hand with the possibility of being the highest hand?

this question i did p(a)=n(a)/n(s)

i subbed in 6 nor n(a), and 20358520 for n(s), and divided but got the answer wrong. I got the value for n(a) wrong, but the n(s) right

Q2:In a horse race you are told that the odds of winning for a certain horse is 3:2 (3wins : 2loses). What is the probability of winning?

this question i have no clue..

Q3: A garage door opener has 10 digits. 4 digits must be punched in sequence in order to open the door. It was noticed that 4 numbers on the pads were worn off more than the others. What is the probability of opening the garage door in one try if you know the 4 numbers but no in order?

for this question is it one divided by 4 factorial?
 
Last edited:

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, chee!

(1) The game of two-handed cribbage is played with a destand deck of 52 cards.
Each player is dealt 6 cards.

(a) How many different six-card hands could be dealt?

For this question I did 52 factorial divided by 46 factorial, but got the wrong answer.
Is it 52! divided by (46!)(61)? . Yes!

Think about it . . .

You had: .\(\displaystyle \dfrac{52!}{46!} \:=\:52\cdot51\cdot50\cdot49\cdot48\cdot47\)

Your reasoning would be:
. . the 1st card can be any of the 52 cards,
. . the 2nd card can be any of the remaining 51 cards,
. . the 3rd card can be any of the remaining 50 cards
. . . . . and so on.

By doing so, you have imparted an order to the cards.
So that: .\(\displaystyle \{A\heartsuit,\,2\heartsuit,\,3\heartsuit,\,4 \heartsuit,\,5\heartsuit,\,6\heartsuit\} \;\ne\;\{2\heartsuit,\,A\heartsuit,\,3\heartsuit, \,4 \heartsuit,\,5\heartsuit,\,6\heartsuit\}\)


The answer is a combination of 52 cards taken 6 at a time:
.. \(\displaystyle _{52}C_6 \;=\;\dfrac{52!}{6!\,46!} \;=\;20,\!358,\!520\)




(b) The highest hand in cribbage has three 5s, one Jack, and two other cards.
The other cards cannnot contain a 5, but can be a Jack.
How many possible highest hands are possible?

There are \(\displaystyle _4C_3 = 4\) ways to get three 5s.

There are \(\displaystyle _4C_1 = 4\) ways to get a Jack.

There are 48 remaining cards , but we must not get the fourth 5.
Hence, there are \(\displaystyle _{47}C_2 \,=\,1081\) ways to get the other 2 cards.


Therefore, the number of highest hands is: .\(\displaystyle 4\cdot4\cdot1081 \:=\:17,\!296\)




(c) What is the probability of being dealt a highest hand?

\(\displaystyle P(\text{highest hand}) \;=\;\dfrac{17,\!296}{20,\!358,\!520} \;=\;\dfrac{46}{54,\!145}\)
 
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