You've got two decks of cards. Each consists of numbers from 1 to 20. After you mixed all, you deal them out to two people. What's the probability that one person has at least one 1.
Solve it!
Jackson Lee
Other urls found in this thread:
en.wikipedia.org
pastebin.com
pastebin.com
twitter.com
Kayden Lewis
>Do my homework hurrr durrr
What's in it for me?
Jace Bailey
2/40+1/39
Jayden Brooks
No, it's not homework, I just want to know, if
66% is right or not.
Preparing for A-level btw
Christian Garcia
Explain, please.
Carson Lopez
the probability of the first person you deal to getting a 1 or the second given the first did not get a 1 is 1/20 (2 1s in the deck out of 40 cards)
^probability that someone gets one 1
the probability of the second person getting a 1 given the first person got a 1 is 1/39 (one 1 out of 39 cards left)
^probabilitynthey both get a 1
add them and that's the probability that at least one 1 shows up (assuming you only deal one card to each person, OP never specified)
Robert Parker
thinking about it now I'm actually wrong
it's 1/39 + 2/40 + (1/39)*(1/20)
Luke Thomas
Alright
P(A or B) = P (A) + P (B) - P(AnB)
P(A or B) = 2/40 + 2/40 - P(A)*P(A/B)
P(A or B) = 4/40 - 2/40 *1/39
P(A or B) = 4/40 - 2/1560
P(A or B) = (156 - 2) / 1560
P(A or B) = 154 / 1560
P(A or B) = 0.0987179487
P(A or B) = ~0.10 or 10%
This is wrong, A and B could happen simultaneously thus P(AnB) =/= 0
Nicholas Barnes
P(only second 1 or only first 1 or both 1)
*all are mutually exclusive therefore
= P(only second 1) + P(only first 1) + P(first 1 and second 1)
= 1/39 (no replacement) + 2/40 + (1/39)*(2/40)
*P(first and second 1)= P(1|one 1)*P(first 1)
^ by rearrangement of the conditional probability formula
Isaac Watson
It's simpler of you think of
A = First person getting a 1
and
B = Second person getting a 1
While P(B/A) is the chance B gets a 1 *after* A has gotten one
B is dependent on A
Neither is mutually exclusive
Thus: