The traditional deck features suits in red (diamonds and hearts) and black (clubs and spades), what is the approximate probability
a. that the card drawn is a seven or a red suit card?
b. of drawing a seven, returning it to the deck and right after pulling a red suit card?
c. of drawing a seven and then a red card without replacing the first card?
a. The probability is given by $\dfrac{4}{52} + \dfrac{26}{52} - \dfrac{2}{52} = 0,53$
b. $\dfrac{1}{13} \cdot \dfrac{1}{2} = \dfrac{1}{26}$
Are my calculations in a and b correct? I can't see any difference between calculating b and c.
b & c are different scenarios, but you get the same answer, still. b states that you need to find the probability of 'drawing a seven, returning it to the deck and right after pulling a red suit card?'. That means you need to find the probability of drawing a 7 from 52 cards, which is $\frac4{52}$. Now, you need to put the card you drew back into the pile, leaving it with 52 cards again. Now find the probability of drawing a red card. There are 2 colors, red and black, each with equal probability, so this is equal to $\frac12$. Now multiply the probabilities you got. $\frac4{52}$ * $\frac12$ = $\frac1{26}$. So that is the answer to b.
c is very similar, but after you draw the seven, you don't put the 7 back in the pile, so there are only 51 cards left. Now we split this up into 2 cases. If the 7 is red, there is a $\frac2{52}$ chance to draw it. Since we drew a red seven, there are 25 reds to draw out of the remaining 51 cards, so the chance of drawing a red card after drawing a red 7 is $\frac{25}{51}$. Now multiply the probabilities $\frac2{52}$ * $\frac{25}{51}$ = $\frac{25}{1326}$. The second case is that the 7 we drew was black, a chance of $\frac2{52}$. This leaves 26 red cards out of 51 left to draw, so the chance of drawing a red is now $\frac{26}{51}$. Multiply the probabilities to get $\frac2{52}$ * $\frac{26}{51}$ = $\frac1{51}$. Now add the probabilities of both cases to get $\frac{25}{1326}$ + $\frac1{51}$ = $\frac{1}{26}$.