1 card of 52 drawn, die rolled. find prob. of red card or 1

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wilber123

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One card is randomly selected from a shuffled standard deck of 52 cards and then a 6 sided die is rolled. Find the probability of getting a red card or a 1.

So far I have

26/52 + 1/6 - x/x = P(red card or 1) I am not sure what goes in the x/x part. I would think my sample space would go on the bottom but how would that be calculated 52 x 6..?

Any help would be greatly appreciated.
 
Re: Flip card and roll a die

probability of a red card= 26/52
let P[R] be probability of a red card
P[R]=1/2
Q[R} BE PROBABILITY OF A NON RED CARD=1/2

probability of a 1 = 1/6
ler P[1] be the probability of a 1
P[1] =1/6
Q[1] be the probability of rolling a non 1=5/6
Q[1]=5/6

there are the following 3 cases that satisfy us"
one red one non 1; or one red one 1 ; or one non red one 1.
P[R]Q[1] or P[R]P[1] or Q[R]P[1]
when we say "or" we add , when we say "and" we multiply

[1/2][5/6] +[1/2][1/6]+[5/6][1/2]
5/12 + 1/12 +5/12
11/12 answer

Arthur
 
Re: Flip card and roll a die

I made an error when I substituted
the answer = 1/12+5/12+1/12
answer = 7/12

another approach is:
probability of at least 1 red or a 1 is
1-the case of non red and no 1

1-Q[R]Q[1]
1-1/2[5/6]
1-5/12
7/12 answer

Sorry for the previous substitution error
 
Re: Flip card and roll a die

Hello Arthur,

I see what you are saying. But if im not mistaking we have to subtract the case where you get a red and roll a 1. My teacher calls it the Union rule.

P(EUF)=P(E)+P(F)-P(EnF)

From what you posted youre saying the sample space is 12 am I correct in that?

Thanks for the help
 
Re: Flip card and roll a die

yes the sample space is 12.
I have trouble indicating it but I will try
...............die
..........1.....2.....3.....4.....5.....6
black
red

sample space 12
P[e]=2/12
P[r]=6/12
p[enr]=1/12

p[e]+p[r]-p[enf]=2/12+6/12-1/12
7/12 answer

I did it slightly differently to obtain 7/12,but "many ways to skin a cat"
Arthur
pr
 
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