, p. 128:8.2.4 A penny, nickel, and dime are simultaneously flipped. What is the probability that heads are showing on at least 6¢ worth of coins?I can do this by brute force, but I don't see the math.
Sunday, April 24, 2011
help desk - probability
from Art of Problem Solving Introduction to Counting and Probability by David Patrick, p. 128:
, p. 128:
Each coin has 2 sides. The total number of combinations is thus 2•2•2=8.
ReplyDeleteOf those 8 combinations, the only ones that would result in less than 6¢ worth of heads would be penny only, and nickel only.
That means that of the 8 combinations, there are only 2 which result in less than 6¢, and 6 that result in at least 6¢.
6/8 = 3/4
The probability is 3/4
Hmmm...the solution manual says 5/8...which is what I get drawing a solution tree----??
ReplyDeleteThere's one more - all three coins show tails.
ReplyDeleteI'd think of it this way: There are 8 total outcomes. Four are heads-up on the dime, so will be over 6 cents no matter what the other two coins do.
So we are only concerned with the four outcomes where the dime is tails. These are:
penny up, nickel up, so 6 cents
penny up, nickel down, so 1 cent
penny down, nickel up, so 5 cents
penny down, nickel down, 0 cents
So 3 outcomes are less than 6 cents. I don't know that you can exactly do this without some brute force.
Penny only
ReplyDeleteNickle only, or
None
5/8
-Mark Roulo
Mathematically, the way I see it is:
ReplyDeleteThe probability of at least 6 cents heads showing is equal to:
a) Penny and Nickel are heads or
b)the Dime is heads
minus
c) probability of D/N/P heads(since this combination is accounted for in the first two calculations)
a) The probability of Penny and Nickel are heads = (.5)*(.5) = .25
or means + in probability space
b) dime is heads = (.5)
minus d/n/p is heads
c) (.5)(.5)(.5)
=
.25 + .5 - .125
= .625
= 5/8
Sorry, I should have said
ReplyDeletethe binomial coefficients are your calculated number of combinations
p (k heads from n independent tosses) = n! / k! x (n-k)!
p (3 heads) = 3! / 3! x (3-3)! = 6/6x1 = 1 (remember 0! = 1)
p (2 heads) = 3! / 2! x (3-2)! = 6/2x1 = 3
p (1 head) = 3! / 1! x (3-1)! = 6/1x2 = 3
p (0 heads) = 3! / 0! x (3-0)! = 6/1x6 = 1
The math in this section was mostly making sure that the students understand mutual exclusivity and don't overcount unintentionally. This problem also makes sure that the student reads critically: "AT LEAST 6 cents" is read as 6 cents or more.
ReplyDeleteEach coin toss is independent of the others. So 2x2x2=8 ways to arrange the 3 coins. Use counting principle, not tree or table b/c later, more complicated problems may be too tedious to write out as tree/table. (If not convinced to leave tree/table, do enough practice to convince yourself that counting principle works and why)
Reason then that all possiblilies that have the dime as heads up will satisfy the condition - that's 1x2x2 or 4 of the 8 combos.
Of the other 4 combos of P and N with D tails, the only way to get six cents or better is nickel heads and penny heads - so just 1 more combo meets the conditions.
Use definition of probability-- P=
desired outcome/possible outcomes-- to find P(Heads showing on at least 6 cents) = (4+1)/8 = 5/8
I discovered a blog by a 2400 SAT Tutor, and he has a post about probability today: http://blog.pwnthesat.com/2011/04/keep-probability-questions-as-simple-as.html
ReplyDeleteI agree with the commenters above that there's no neat and tidy equation that'll get you there, since the event you seek is itself defined by a mathematical relationship. I would just list the possibilities (using the values for the coins):
ReplyDelete0+0+0 = 0
0+0+1 = 1
0+5+0 = 5
0+5+1 = 6
10+0+0 = 10
10+0+1 = 11
10+5+0 = 15
10+5+1 = 16
5/8 of the possibilities add up to at least 6 cents.
THANK YOU!!!
ReplyDeleteSOOOOO helpful!
Trying to teach yourself counting & probability from a book is crazy (I think).
I've reached the point where I REALLY need a teacher.
Not just a book.
a) Penny and Nickel are heads or
ReplyDeleteb)the Dime is heads
minus
c) probability of D/N/P heads(since this combination is accounted for in the first two calculations)
I'm proud to say that at some point, as I was obsessing over this problem, I figured this out.
I still don't understand the solution manual's solution - must post what they say.
All of your solutions are MUCH more clear.