What is the probability of picking a red card or a 7, if 1 card is sel

Probability of getting a red card: 26/52
Probability of getting a 7: 4/52
Since we have "or", we add these and get 30/52 or 15/26

Could someone please explain where am I going wrong?

P{R u 7} = P{R} + P{7} - P{R n 7} = (26 + 4 - 2)/52 = 28/52 = 7/13

Hey csaluja as I have mentioned above in my post, there are a total of 26 red cards in a deck and a total of 4 cards of 7. 2 of the 7s are red. So you get \(26+2=28\)

And thus,

\(\frac{28}{52}\)
\(=\frac{14}{26}\)
\(=\frac{7}{13}\)

Hope that helps.

Hey csaluja

I am happy to help!

As you have already mentioned correctly, the probability of getting a 7 is 4/52.
Also, the probability of receiving red is exactly 26/52.
So let us start from here.

If you think about the question, what it really asks is, when will we receive either red or 7.
So the two events are not mutually exclusive! We know that there are certain cases in which the 7 could be drawn and show red! Well, how many cases where this happens are there? Out of 4 cases, two such cases ( (1/2) *(4/52) = 2/52) could occur! ( These two cases should be subtracted from the total cases in which we draw the 7 AND/OR a red card.

Let us do the math:

P (red OR #7) = P(red) + P(#7) - P(red*#7)

P (red OR #7) = (26/52) + (4/52) - (2/52) = (26/52) + (2/52) = 7/13

Thus, the OA is 7/13

Best, gota900

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