HW2
assignment
1.
2.
3.
A fair coin is tossed 5 times. Find a probability of obtaining (a) exactly 3
heads; (b) at most 3 heads
Solution:
Binomial distribution.
(a) P(n=3) = C5,3 0.530.52 = Binomial[5,3]*0.55 = 0.31
(b) P(n<=3) = 1-P(4)-P(5)= 0.81
A hat contains tags numbered 1,2,3,4 and 5. A tag is drawn from the hat
and replaced, then the second tag is drawn. (a) show the sample space;
(b) Find the probability that the number on the second tag exceeds the
number on the first.
Solution:
(a) S includes 25 points, {x,y}, Similar to a sample space for two dice with 5
faces each
(b) There are 10 points in S where y>x. P(y>x) = 10/25 = 2/5.
(c) Find the probability that the first tag has a prime number and the second tag
has an even number (1 is not considered a prime number)
Solution: The points are : {22},{24},{32},{34},{52},{54}. P=6/25 = 0.24.
There are 4 roads connecting the cities A and B and 3 roads between the
cities B and C. How many ways can round trip be made from city A to C
(through B) and back? How many ways if it is desired to take different
routs on the way back?
4.
(a) How many license plates are possible if the first three places are occupied
by letters and the last three by numbers
(b) Assuming all combinations are equally likely, what is the probability the three
letters and the three numbers are different?
Solution
(a) 263 103
(b) P26,3P10,3/ 263 103 = 0.64
5.
Six students, three boys and three girls, line up in a random order for a
photograph. What’s the probability that the boys and girls alternate?
Solution
N= 3!*3! + 3!*3!= 72; P = 72/6!= 72/720= 0.1
6. Ten children are to be grouped into two clubs (Lions and Tigers)
with 5 children in each club. Each club is then to elect a
president and a secretary. In how many ways can this be done?
Solution
C10,5P5,2P5,2=100800.
7. N people are arranged in a line. What is the probability that 2
particular people, say A and B, are not next to each other?
Solution:
Consider two people as a unit. There are (N-1)! Ways to
permute such a group. The total number of permutations
corresponding to A and B placed together is 2*(N-1)!.
The probability P(A and B are together) = 2(N-1)!/N! = 2/N.
The answer is P(A and B separated) = 1 – 2/N = (N-2)/N
8. How many ways can 4 couples sit at a round table if each man sits next
to two women neither one of whom is his wife
Placing men in every other chair (even or odd) results in 2 4!
possibilities.
Thus, we have 48 possible arrangements.
For each of these arrangement, for every empty chair there are two
possible choices of a woman per such that she is married to none of
her two neighbors.
Solution
(1) Let’s place men first. The two possibilities are (a) every other odd
chair; (b) every other even chair. The total number of arrangements is
2*4! = 48; (2) The second step is to place the women in the empty
chairs. For each placement of the men, there are two possible
arrangements of the women that satisfy the conditions.
Consider an example:
{M1 W3 M2 W4 M3 W1 M4 W2}
{M1 W4 M2 W1 M3 W2 M4 W3}
The answer is N=48*2 = 96.
9. How many times should you roll a die so that the probability of at least
one 6 is at least 0.9?
Solution
P(m,n) is the probability of m six’s in n rolls.
P(m>=1,n) =1-P(m=0,n);
P(0,n) = (5/6)n
This probability must be less or equal to 0.1 (so that the probability of
the compliment event, “at least one 6”, is >= 0.9.
Thus we have to find an integer n with which P(0,n) is the number closest
to 0.1 from below.
We can first solve the equation
(5/6)n = 0.1;
n = Log[0.1]/Log[5/6]= 12.6;
The closest integer solution is 13.
You can also do it with Mathematica by making a table of P(0,n) for
{n,0,13}
10. A student has no knowledge whatsoever of the material to be tested
on a true-false examination, and so the student flips the coin in order
to determine the response to each question. What is the probability
that the student scores at least 60% on a ten-item examination?
Solution:
“At least 6” means : 6 OR 7 OR 8 OR 9 OR 10. The events are disjoined
and the probability equals the sum of the individual probabilities:
0.510 (Binomial[10,6]+Binomila[10,7]+… +Binomial[10,10])=0.377
11. If 70% of the bolts produced by a machine are perfect,
20 % are of average quality and 10 % are defective, determine
the probability that out of 15 bolts chosen at random, (a) exactly
two are average, (b) 10 are perfect, 3 are average and 2 are
defective, (c) less than 3 are defective. 11.
Solution
(a) Exactly two are “average”.
In this case the problem is reduced to Binomial distribution with m=2,
p=0.2,q=0.8
P(2 average) = Binomial[2,15] (0.2)^2 * (0.8)^13= 0.23
(b) 10 are perfect, 3 are average and 2 are defective.
Now we have three outcomes, and have to apply a multinomial
distribution
P(10,3,2) = (15!/10!3!2!) 0.7100.230.12 = 0.07
(c) Less than 3 are defective.
Binomial, with p =0.1 and q = 0.9.
P(ndef<3) = P(0)+P(1)+P(2) = Binomial[15,0]0.915 +
Binomial[15,1]0.9140.1+ Binomial[15,2]0.9130.12 =0.82