Question

ECS20: Discrete Mathematics UC Davis Instructor: Dale Fletter Student Name Here!!! Problem Set 3 Due September 4, 2019, noon in homebox located 2131 Kemper Problem 1: (1 points) In the card game of poker, five cards that have consecutive kinds is called a straight An ace can be considered either the lowest card of an A-2-3-4-5 straight or the highest card of a 10 J-Q-K-A straight. What is the probability that a five-card poker hand contains a straight? For this problem, you may consider a suited straight (all cards of the same suit) as a straight although the rules of poker exclude that as a straight Problem 2: (1 points) In the questions below an experiment consists of randomly choosing a bit-string of length five. Consider the following events: E1: the bit string choscn begins with 1 E2: the bit string chosen ends with 1 E3: the bit string chosen has exactly three 1s. (a) Find p(E1E3) (b) Find p(F32) (c) Find p(E2E3) (d) Find p(E3 E1n E2) (e) Determine whether El and E2 are independent. (f) Determine whehter E2 and E3 are independent Problem 31 points) Prove that 33.5352 ... +3-5" = 3(5"+-1)/4 whenever n is a nonnegative integer Page 1 of

Answer 1

Problem 2:

Total number of possible bit strings of length 5 = 2⁵ = 32  

First, we need to calculate the probabilities of events E1, E2, and E3.

E1: the bit string chosen starts with 1.

This means that the first bit must be 1 and the rest of the 4 bits can be anything.

So, for the first bit, we have only one choice (i.e., 1), but for each of the remaining four bits, we have 2 choices each (i.e., 0 or 1).

1 1 x 2 x 2x 2x 2 16

Thus, P(E1) = (1x2x2x2x2) / 32 = 16 / 32

E2: the bit string chosen ends with 1.

This means that the last (fifth) bit must be 1 and the rest of the 4 bits can be anything.

So, for the first four bits, we have 2 choices each (i.e., 0 or 1), but for the fifth bit, we have only one choice (i.e., 1).

2x 2x 2X 2X 1 16

Thus, P(E2) = (2x2x2x2x1) / 32 = 16 / 32

E3: the bit string chosen has exactly three 1s.

In this case, the bit string chosen should have exactly three 1s. That means the other two bits will always be 0. So now, we need to choose 3 out of 5 positions for 1 and we don't care about the permutation of 1s among themselves since all the 1s appear same. Therefore, we will use combination here.

Thus, P(E3) = ⁵C₃ / 32 = 10 / 32

Now, we need to calculate P(E1∩E2), P(E2∩E3), and P(E1∩E3).

E1∩E2

This means that the string both starts and ends with 1. This means that for the first and last position, we have only one choice (i.e. 1), but for the remaining 3 positions. we have 2 choices each (0 or 1).

Thus, P(E1∩E2) = (1x2x2x2x1) / 32 = 8 / 32

E2∩E3

This means that the last bit will always be 1 and the string contains exactly three 1s. This means that the first four bits will have exactly two 1s. Again, we need to apply combination to choose two positions out of four for 1.

Thus, P(E2∩E3) = ⁴C₂ / 32 = 6 / 32

E1∩E3

This means that the first bit will always be 1 and the string contains exactly three 1s. This means that the last four bits will have exactly two 1s. Again, we need to apply combination to choose two positions out of four for 1.

Thus, P(E1∩E3) = ⁴C₂ / 32 = 6 / 32

E1∩E2∩E3

This means that the first and last bit will always be 1 and the string contains exactly three 1s. This means that the remaining 3 bits will always contain exactly one 1 and two 0s. Thus, we again need to apply combination to choose one position out for three for 1.

Thus, P(E1∩E2∩E3) = ³C₁ / 32 = 3 / 32

Now, we have all the probabilities. Lets now use them to find the solutions to the given problem.

(a). P(E1 | E3) = P(E1∩E3) / P(E3)

= (6 / 32) / (10 / 32)

= 3 / 5 = 0.6

(b). P(E3 | E2) = P(E2∩E3) / P(E2)

= (6 / 32) / (16 / 32)

= 3 / 8 = 0.375

(c). P(E2 | E3) = P(E2∩E3) / P(E3)

= (6 / 32) / (10 / 32)

= 6 / 10 = 0.6

(d). P(E3 | E1∩E2) = P(E1∩E2∩E3) / P(E1∩E2)

= (3 / 32) / (8 / 32)

= (3 / 8) = 0.375

(e). Two events E1 and E2 are independent if P(E1∩E2) = P(E1).P(E2)

P(E1∩E2) = 8 / 32 = 0.25

P(E1).P(E2) = (16/32).(16/32) = 0.25

This means that P(E1∩E2) = P(E1).P(E2)

Hence, E1 and E2 are independent.

(f). Two events E1 and E2 are independent if P(E1∩E2) = P(E1).P(E2)

P(E2∩E3) = 6 / 32 = 0.1875

P(E2).P(E3) = (16/32).(10/32) = 0.15625

This means that P(E2∩E3) ≠ P(E2).P(E3)

Hence, E2 and E3 are not independent.

Problem 3:

The following figure shows how to solve this problem:

3t 3.5 t 3.5+ 3.5h =3 Cit 5 + 52+ - -- + 5") 3 (5°+ 5't + 5) 52 t GP with a51 f terms n no. (5-1 3X ntt 5-1 3 (5-1

First, we took out 3 common from the given series. Then, the resultant series is a Geometric Progression with a=1, r=5, and the number of terms = n+1. We then apply the formula for the sum of a GP and the result is the same as the RHS given in the question. Hence proved.

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