Midterm 1 Solution
1) An experiment consists of rolling two dice simultaneously.
a) Find the sample space S (3 pts)
The sample space is the set of 36 events as follow
S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }
b) What is the probability that you didn’t get a six for either dice until the third roll?(5pts)
P{Not getting any six until the 3rd roll}= ,
where p is the probability of getting a six on either dice. That is
.
Thus,
P{Not getting any six until the 3rd roll}=
c) What is the probability that, in 5 rolls, there are exactly 2 instances in which, at least one of the dice is a six? (5pts)
P{ 2 of 5 rolls has at least one of the dice is a six}
=
, where p is the probability that one of the dice is a six. That is, p = .
Therefore, P{ 2 of 5 rolls has at least one dice is a six}=
2) A digital communication system transmits one of the three values -1, 0, 1. Due to impairments on the channel, the receiver sometimes makes an error. The error rates are 12.5% if -1 is transmitted, 75% if 0 is transmitted, and 12.5% if 1 is transmitted. If the probabilities for the symbols 1 and -1 being transmitted 1/4,
a) Find the probability of 0 being transmitted (2pts)
P{0 being transmitted}= =
b) Find the probability of error (10pts)
P{error}= P{not receiving the same value as transmit}
= P{error|-1 is transmitted} P{-1 is transmitted}+
P{error| 0 is transmitted} P{ 0 is transmitted}+
P{error| 1 is transmitted} P{ 1 is transmitted}
= (1/8)* (1/4) + (3/4)* (1/2) + (1/8)* (1/4)
= (1/32) + (3/8) + (1/32)
= (7/16)
3) A random variable X has PMF
a) What is the value of the constant c? (5pts).
Since = = 1,
and
,
thus,
.
b) What is (5pts)
4) Bonus Problem: (5pts)
Three con so that he or she can't see it. The color of each hat is based on a coin Three contestants, Alice, Bob, and Cedric, enter a room and a hat is placed on each contestant's head so that he or she can't see it. The color of each hat is based on a coin toss – blue (B) for heads, red (R) for tails. After all the contestants enter the room, they look at the colors of the others' hats and, based on this information, they guess the color of their hat. They can guess red or blue, or, if they can't make up their mind, they can pass. No communication is allowed during the competition, but the players are allowed to agree on a strategy before play begins. The team wins if at least one of them guesses correctly, and none of them guesses incorrectly. What is the team's best strategy?
Solve
We know,
P(0 out of 3 is R) = P(3 out of 3 is B)
P(1 out of 3 is R) = P(2 out of 3 is B)
P(2 out of 3 is R) = P(1 out of 3 is B)
P(3 out of 3 is R) = P(3 out of 3 is B)
Thus, the probability that 2 out of 3 is Red, or 2 out of 3 is blue is higher than other events. Thus, we make decision based on “2 out of 3 strategy” as
“Guess from what we see. If we see other hats are both red/ blue, we guess the opposite one. If they have different color, we pass.”
P(win) = 1-P(guess wrong)
= 1-P(at least one person gets wrong)
= 1-P(comes out all blue or all red)
= 1-(1/8+1/8)
= 6/8