SOLUTION: Can someone please with this one. A board game uses the deck of 20 cards shown. 4 rows of cards and they are from 1-5. firt row - 1,2,3,4,5 second row - 1,2,3,4,5 thrid ro

Algebra ->  Probability-and-statistics -> SOLUTION: Can someone please with this one. A board game uses the deck of 20 cards shown. 4 rows of cards and they are from 1-5. firt row - 1,2,3,4,5 second row - 1,2,3,4,5 thrid ro      Log On


   



Question 469398: Can someone please with this one.
A board game uses the deck of 20 cards shown. 4 rows of cards and they are from 1-5.

firt row - 1,2,3,4,5
second row - 1,2,3,4,5
thrid row - 1,2,3,4,5
fourth row - 1,2,3,4,5
Two cards are selected at random from this deck. Determine the probability of they both show even numbers
a) with replacement
b) without replacement

Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Given 4 sets of {1,2,3,4,5}
Notice there are 3 odd (1,3,5) and 2 even (2,4)
Multiply by 4, which gives 12 odds and 8 evens in entire deck
So chance of picking an even number is 8/20= 2/5
With replacement:
Each selection is independent in this case, so probabilities remain constant
Probability of picking even on 1st card is 2/5
Probability of picking even on 2nd card is 2/5
%282%2F5%29+%2A+%282%2F5%29+=+4%2F25
Therefore probability of both cards being even is 4/25 or 0.16
Without replacement:
selections are not independent of each other, so probabilities may change
Probability of picking even on 1st card is 2/5
Now there is one less card in deck and one less even card
total = 19, even = 7, odd = 12
Probability of picking even on 2nd card is 7/19
%282%2F5%29+%2A+%287%2F19%29+=+14%2F95
Therefore probability of both cards being even is 14/95 or 0.147