Lauren's thoughts

OCCUPATION: PhD student FAVORITES: being with my wonderful husband Steve, stand-up comedy, travel, Harry Potter, reading for enjoyment, music (eclectic: orchestral, jazz & blues, top 40, oldies), cooking vegetarian cuisine, spending time with friends, watching "my shows" (Grey's Anatomy, Lost, 24, Six Feet Under, Sopranos) & shopping at Lerner NY & Co! DISLIKES: accidentally eating meat, hatred, intolerance, odd numbers, & dirty kitchens

Tuesday, July 12, 2005

a mind teaser

I used this in my class today (which by the way, is going great!). I won't say HOW or WHY I used it, so that if you want to give it a shot, you can start from scratch. Try it out and post a comment with your response and reason why...

There are two bags: one bag (R) contains exactly 100 red jelly beans and the other bag (B) contains exactly 100 black jelly beans. First, 15 random red jelly beans are counted out and poured into bag B (the one that contains all black jelly beans). Then, that bag is shaken to mix up the colors. Then, any 15 random jelly beans from bag B (the one that contains 100 black jelly beans and 15 red), are counted out and poured into bag R (the one that contains the red jelly beans).

TARGET QUESTION: “Will the number of red jelly beans in bag B (that initally contained only black jelly beans) be the same as the number of black jelly beans in bag R (that initially only contained red jelly beans)?

2 Comments:

Blogger erica said...

Okay, well, I'll start off. Yep, it's the same number. Say your 15 from B happens to be 12 black and 3 red that you then put back into R. The key is that you'd be leaving behind however many reds you didn't take to put back into R, which in this case would be 12, so the B bag would then have 12 red and the R bag would have 12 black. It works for all possibilities.

I'll take the B bag, because I like licorice jellybeans and I hate red icing. =)

Jack would like this, since he's a future statistician!

2:36 PM  
Blogger Lauren said...

Is this thing on? It says 0 comments, but when I went to post one to see if it is working, Erica's showed up. Well here's what she said:

Okay, well, I'll start off. Yep, it's the same number. Say your 15 from B happens to be 12 black and 3 red that you then put back into R. The key is that you'd be leaving behind however many reds you didn't take to put back into R, which in this case would be 12, so the B bag would then have 12 red and the R bag would have 12 black. It works for all possibilities.

I'll take the B bag, because I like licorice jellybeans and I hate red icing. =)

Jack would like this, since he's a future statistician!

9:55 AM  

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