Physics 6
Spring 2005
Some Practice Problems for Exam 2
SOLUTIONS
1. What is the probability of getting
only
2 heads in 4 tosses of a coin?
There are 4!/(2!2!)=6 ways to have 2 heads distributed among 4 tosses. You
do not need the formula for this; just count the possibilities (HHTT, HTHT,
HTTH, THHT, THTH, TTHH). The probability for 2 heads and 2 tails (in any
order) is (1/2)
2
(1/2)
2
for one of those orderings. So the overall probability is
6/(2)
4
=6/16=3/8=0.375
2. What is the probability of getting
at least
2 heads in 4 tosses of a coin?
Add the prob for 2 heads in 4 tosses (from above) to prob for 3 heads in 4
tosses (4!/3!=4 orderings times (1/2)
3
(1/2)
1
or 4/16=1/4) to prob for 4 heads in
4 tosses (1/16). Get 3/8+1/4+1/16=11/16=0.6875
3. Consider
4-sided dice
.
a. What is the probability of getting
exactly
2 number 3's in 4 tosses?
6 orderings (as in (1) above) times (1/4)
2
for two number 3's and (3/4)
2
for
two not number 3's OR 6(1/4)
2
(3/4)
2
=54/256=0.211
b. What is the probability of getting
at least
2 number 3's in 4 tosses?

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