Quiz 4 solutions

 

If you have any questions regarding the solutions mail to  Sreenivas Rao Kollu

 

1. A pair of fair dice is rolled once. What is the probability that the total number of points on the dice is eight? ( 3 points )
(4,4),  (5,3),  (3,5),  (6,2),  (2,6)  five favourable cases/outcomes.
Total number of cases/outcomes = 6 x 6 = 36.
Probability = (number of favourable outcomes)/ (total number of outcomes) = 5/36.

 

2. A fair coin is tossed 1000 times.
(a) What is the probability of observing exactly 500 heads? ( 2 points )
Bernoulli Trials
Out of 1000 tosses  500 must be heads and (1000 – 500) = 500 must be tails.
choosing 500 tosses out of 1000 tosses – C(1000, 500)
and (probability of  observing head) x ( probability of observing head) .. 500 times X
(probability of observing tail) x (probability of observing tail) … 500 times
The tossed coin is fair. So probaility of observing head = probability of observing tail = 1/2.
The probability of observing 500 heads out of 1000 tosses
            =   C(1000, 500) x (1/2)500 x (1/2)500    =   C(1000, 500) (1/2)1000

(b) Write an expression for the probability of observing a number of heads between 450 and 550 inclusively? ( 2 points )
Bernoulli Trials
Total number of tosses between 450 and 550 inclusively = 101.
The probability of observing  N number of heads and (101 - N) number of trial in those 101 trials
             = C(101, N)  x (1/2)N  x (1/2)101-N  =  C(101, N) (1/2)101

Well this is one possible way of understanding the question (i.e. N heads between the 450-th and 550-th toss). What I had originally meant

is the probability of obtaining a total number of heads between 450 and 550 over the 1000 tosses. This probability is just the sum of the probability

of all the disjoint event associated with getting 450 heads, or 451, or 452, etc. In this case the answer is best written as:

C(1000,450) (1/2))450(1/2)550  + C(1000,451) (1/2))451(1/2)549  + ... + C(1000,550) (1/2))550(1/2)450

which can be further simplified by using the "sigma" notation and collecting all the terms containing (1/2)1000. (PB)

 

3. What is the probability that a five- card  poker hand consists of five black cards? ( 3 points )
There are 26 black cards.
Number of favourable cases = choosing 5 cards out of those 26 black cards = C(26, 5).
Total Number of cases = choosing 5 cards out of 52 cards = C(52, 5).
Requiredprobability = C(26, 5) / C(52, 5).

 

If you have any questions regarding the solutions mail to  Sreenivas Rao Kollu