kitchen table math, the sequel: help desk - probability

Sunday, April 24, 2011

help desk - probability

from Art of Problem Solving Introduction to Counting and Probability by David Patrick, 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.


Obi-Wandreas, The Funky Viking said...

Each coin has 2 sides. The total number of combinations is thus 2•2•2=8.

Of 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

Catherine Johnson said...

Hmmm...the solution manual says 5/8...which is what I get drawing a solution tree----??

ChemProf said...

There's one more - all three coins show tails.

I'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.

Anonymous said...

Penny only

Nickle only, or



-Mark Roulo

Anonymous said...

Mathematically, the way I see it is:

The probability of at least 6 cents heads showing is equal to:

a) Penny and Nickel are heads or
b)the Dime is heads
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

Anonymous said...

Sorry, I should have said

the 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

lgm said...

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.

Each 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

debbie stier said...

I discovered a blog by a 2400 SAT Tutor, and he has a post about probability today:

PWN the SAT said...

I 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):

0+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.

Catherine Johnson said...


SOOOOO 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.

Catherine Johnson said...

a) Penny and Nickel are heads or
b)the Dime is heads
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.