1. A far die is rolled five times. a. What is the probability of getting five 6?

Question

1. A far die is rolled five times. a. What is the probability of getting five 6?s in a row? b. What is the probability of getting all even numbers? 2. Network Reliability: In the network shown below, the links between computers can be either up or down. The reliability of each link is as shown in the diagram. The status of each link is independent of the status of the other links. What is the probability that there is a functioning set of links connecting computer B to computer C? Uptime = 3. Consider drawing one card from a standard52-card deck. Define four events as follows: A = {Hearts}, B = {Black cards}, C = {Aces}, and D = {Queen of Hearts, Queen of Spades). Prove which pairs of these events are independent and which are dependent. 4. An urn contains 7 red balls and 5 green balls. Two balls are drawn at random from the urn. What is the probability of getting two balls of the same color if: 5. Communication Channel Reliability: Consider a communication channel over which a 0 or a 1 must be transmitted. Suppose that the probability that the bit to be sent is a 0 is 0.5, and the probability that a 1 is to be sent is also 0.5. Also suppose that due to noise the probability that a 0 is changed to a 1 during transmission is 0.06 and the probability that a 1 is changed to a 0 is 0.03. Suppose that a 0 is received. What is the probability that a 0 was sent?

Explanation / Answer

a) P(five 6's in a row)

So, on each roll, a 6 must pop up

So, on a single roll, any of 1,2,3,4,5 or 6 can pop up

So, Probability that a 6 pops up on a single row is : 1/6

So, P(five 6's in 5 rolls) = (1/6) * (1/6) * (1/6) * (1/6) * (1/6) ----> 1/6^5 ----> ANSWER

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b) P(5 even numbers on 5 rolls) :

So, on each roll, any of 2,4 or 6 can pop up

So, P(even number on a single roll) = 3 / 6 ---> 1/2

So, P(5 even numbers) --> (1/2)^5 ---> 1/2^5 ---> 1/32 ---> ANSWER

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2)

B to C can be either :
a) B to C direct
OR
b) B to A and A to C

So, for B and C to not be connected, we can have any of these scenarios :

Only BA connected
OR
Only AC connected

P(only BA working) = 0.85 * 0.04 * 0.07 = 0.00238
P(only AC working) = 0.96 * 0.15 * 0.07 = 0.01008

So, P(B not connected to C) = 0.00238 + 0.01008 ---> 0.01246

So, P(B connected to C) = 1 - 0.01246 ----> 0.98754 ----> ANSWER

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3)

A = Hearts
B = Black cards
C = Aces
D = Queen of Hearts , Queen of Spades

A and B :
A heart can never be a black card
So, A and B are INDEPENDENT

B and C :
A black card can be an ace because we have Ace of spades and Ace of Clover
So, B and C are DEPENDENT

C and D :
C are aces
D are queens of H and S
A queen can never be an ace
So, C and D are INDEPENDENT

A and C :
A heart can be an ace, i.e the ace of hearts
So, A and C are DEPENDENT

A and D :
A = hearts
D = Queen of H , Queen of Spades
The queen of hearts is common to both A and D
So, A and D are DEPENDENT

B and D :
B = black cards
D = queen of Hearts , queen of Spades
q of Spades is a black card
So, B and D have an element in common
So, B and D are DEPENDENT

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4)

Red = 7
Green = 5

a) with replacement :
Since we are replacing, the second pick is the exact same as the first pick
So, we need two reds or two greens
P(2 reds) = (7/12)^2
P(2 greens) = (5/12)^2

So, P(2 balls of same color) = (7/12)^2 + (5/12)^2 --> (49 + 25) / 144 --> 74/144 ---> 37/72 --> ANSWER

b) without replacement :
P(2 reds) = (7/12)*(6/11)
P(2 greens) = (5/12)*(4/11)

So, P(2 balls of same color) = (7/12)(6/11) + (5/12)(4/11) ---> 31/66 ---> ANSWER

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5)

P(zero sent) = 0.5
P(one sent) = 0.5
P(zero sent but changed to one) = 0.06
P(one sent but changed to zero) = 0.03

P(zero sent) = 0.5 - 0.06

P(zero sent) = 0.44

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