There are two boxes - one holds 11 cards numbered 1 through 11, the other 5 cards numbered 1 through 5. Two cards are randomly drawn. What is the probability of drawing 2 even cards in a row?

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mixtyggc · Accepted Answer

Step 1The first thing we have is 2 boxes, which I&#039;ll call Box A (which holds 11 cards) and Box B (which holds 5 cards).P(Box A)=P(Box B)=12Draw 1, Box A, chance of drawing an even cardLet&#039;s say we draw from Box A. We have cards numbered 1-11, and of the 11 cards, 5 are even. And so we can say that if we draw from Box A, the probability of drawing an even card is 511Draw 1, Box B, chance of drawing an even cardBut what if instead we draw from Box B. We have cards numbered 1-5, and of the 5 cards, 2 are even. And so we can say that if we draw from Box B, the probability of drawing an even card is 25Draw 1, probability of drawing an even cardWe can now put the above together to calculate the probability of drawing an even card:P(drawing an even card)=12×511+12×25=522+15⇒⇒522(55)+15(2222)=25110+22110=47110⇒522(55)+15(2222)=25110+22110=47110We can now move on to the second draw.Draw 2, Box A (where draw 1 was also Box A)In this possibility, we drew from Box A in draw 1 and are doing so again in draw 2.There are now 10 cards and 4 of them are even, and so we get 410=25Draw 2, Box A (where draw 1 was Box B)In this possibility, since we drew from Box B before, Box A still has 11 cards and 5 evens, and so that&#039;s 511Draw 2, Box B (where draw 1 was also Box B)In this possibility, we drew from Box B in draw 1 and are doing so again in draw 2. There are now 4 cards and 1 of them is even, and so we get 14Draw 2, Box B (where draw 1 was Box A)In this possibility, since we drew from Box A before, Box B still has 5 cards and 2 evens, and so that&#039;s 25Draw 2, probability of drawing an even cardAll four of the above possibilities is equally likely, and so we&#039;ll multiply each of the above probabilities by 14 and then sum them up:P(drawing an even card on second draw)=14×25+14×511+14×25+14×14=110+544+110+116⇒110(8888)+544(2020)+110(8888)+116(5555)88880+100880+88880+55880=331880And now we can finalize this calculation by multiplying in the probability of getting an even card on the first draw:

There are two boxes - one holds 11 cards numbered

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