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A Bingo card is a 5x5 array with 24 numbers chosen from 1 to 75 with a "free space" in the middle. Numbers are called in random order, and a player wins when he gets five in a row (horizontally, vertically, or diagonally).

Corey is playing Bingo and notices that she won when the 5th number is called (so she could have won on the 4th or 5th number called). What is the probability that her winning line included the free space?

 Mar 22, 2020
 #1
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0

there are 5 ways to bingo across      1 has free space

2 diagonal ways    both have free spce

5 vertical ways....one has free space

 

 

 

4 with space   out of  17       4/17

 Mar 22, 2020
 #2
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+1

Sorry  that was    4 ways with free space    out of 12 ways    4/12 =   1/3

 Mar 22, 2020
 #3
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Hey guest! Glad to see you trying the problem. I skimmed it over though, and it appears it isn't as simple as finding how many ways there are to win. That's because for all we know, Corey could've won on her 11th or 12th turn(I don't know the exact number). The point is, we don't know on what turn she won on, meaning the cases are a lot more varied than you previously accounted for. 

jfan17  Mar 22, 2020
 #4
avatar+1956 
+1

Nice, I agree!

CalTheGreat  Mar 22, 2020
 #5
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+1

But, the problems says that she won on the fifth number; so, it doesn't go beyond that number.

 

Also, no matter how many numbers are called, only 4 out of 12 ways to win include the free space.

 

I believe that the original answer, 1/3, is correct.

 Mar 22, 2020
 #6
avatar+483 
+3

I got an answer for you now that I've thought about it:

 

Realize that there are a total of 8 ways of winning that DON'T include the free space, and for each of these 8 ways, there's only one possible combination of these 5 numbers that let's corey fill it. There are 4 winning lines that DO include the free space, and each has 71 combinations of called numbers because it's equivalent to the four numbers in the non free spaces plus any of the leftover. The cumulative probability of corey's winning line also including the free space in the middle is equivalent to:

\(4*71/(8+4*71) = 71/73\)

.
 Mar 22, 2020
 #7
avatar+531 
+2

Thanks for the help! I got the same answer as you, but I thought I was definitely wrong. Your explanation makes it a lot more clear. 

MathCuber  Mar 22, 2020
 #8
avatar+483 
+2

No problem!

jfan17  Mar 22, 2020

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