We start with 6 Black playing cards and we add cards to the point where the probability of drawing 2 Red cards in a row without replacement is #1/2#. How many cards do we add?

Question

Julia Adams · Accepted Answer

#15# red cards.

Let the number of red cards to be added to #6# black cards be #x#, where #x# is natural number. Now probability of getting first red card is #x/(x+6)# and conditional probability of getting second red card (assuming we have already got one red card) is #(x-1)/(x+5)#, as we are considering without replacement. Hence probability of drawing #2# red cards, without replacement, is #x/(x+6)xx(x-1)/(x+5)# and as this probability is #1/2#, we have #(x(x-1))/((x+6)(x+5))=1/2# or #2x(x-1)=(x+6)(x+5)# or #2x^2-2x=x^2+11x+30# or #x^2-13x-30=0# or #(x-15)(x+2)=0# i.e. #x=15# or #x=-2# But as number of red cards can only be a natural number #x=15#

Abigail Allen · Answer

If we only add Red cards, 15. If we don't limit the colour of cards added, there is no number that can be added that will achieve that probability.

Let me first point out that if we're talking about solely red cards being added, the answer is 15 and the details are here: https://tutor.hix.ai That said, this question does not limit itself to red cards being added, and so we can have both black and red added. If we don't limit ourselves to a single pack of cards and can instead draw from an infinite supply of mixed cards , we can approach the question this way - first I'll lay out the starting argument the same way the question handling adding solely red cards does: We have (B)lack cards and (R)ed cards. We start with #B=6# and want to add R so that the odds of drawing 2 R, without replacement, is #1/2#. How many R must we add? The odds of drawing an R on a single draw is: #R/(R+B)# and so for example if we have #R=B=6#, we'd have the odds of drawing an R to be: #6/(6+6)=6/12=1/2# So now let's add that second draw into the mix. We've already drawn an R and so we have one R less, so the ratio for the second draw is: #(R-1)/((R-1)+B)# Which means that the two draws taken together are: #(R/(R+B))((R-1)/((R-1)+B))# Now in the analysis with working with only adding R, we know that #B=6#. We don't know that here. We know that #B >=6# and that, in general, the number of R we add will be equal to the number of B added, so we can express that as: #B=6+R# We still want the odds of the two draws to be #1/2#, so we'll get: #(R/(R+6+R))((R-1)/((R-1)+6+R))=1/2# #(R/(2R+6))((R-1)/(2R+5))=1/2# #(R(R-1))/((2R+6)(2R+5))=1/2# #(R^2-R)/(4R^2+22R+30)=1/2# #2(R^2-R)=4R^2+22R+30# #2R^2-2R=4R^2+22R+30# #0=2R^2+24R+30# And now I'll use the Quadratic Formula: # x = (-b \pm sqrt(b^2-4ac)) / (2a) # # x = (-24 \pm sqrt(24^2-4(2)(30))) / (2(2)) # # x = (-24 \pm sqrt(576-240)) / 4 # # x = (-24 \pm sqrt(336)) / 4 # # x~= (-24 \pm 18.33) / 4 # And all we end up with is negative results. This means there are no number of mixed cards that will give us the probability of drawing 2 R in a row with the odds of #1/2#.

Jameson Allen · Answer

undefined You would need to add 7 additional cards to the 6 black playing cards to reach the point where the probability of drawing 2 Red cards in a row without replacement is 1/2.