What is the probability that Mark, Colleen. Jett, and Emily win the first, secon

Question

What is the probability that Mark, Colleen. Jett, and Emily win the first, second, third, and fourth prizes, respectively, in a drawing if 30 people enter a contest and no one can w in more than one prize. winning more than one prize is allowed. To roll a total of 8. do you prefer two dice or three dice Justify your answer. Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit arc equally likely. the probability that a bit is a 1 is 0.7. the probability that the ith bit is a 1 is 1/2^i = 1. 2, 3 Find the probability that a randomly generated bit string of length 10 begins with a I or ends with a (X) for the same conditions as in parts (a), (b). and (c) of (i). if bits arc generated independently.

Explanation / Answer

Answer:

1)
Assumptions:
## All 30 players are equally likely to secure any position.
## No two or more players secure the same position(it is different from one player securing two or more positions & I don't know is it okay to make this assumption or not).
a) In this case, winning the first prize, winning the second prize etc are not independent events because if a person wins a prize he/she can not win the other prizes.
As I have already assumed that no two or more players secure the same position.So we have 30 different positions to be secured by 30 people.
All possible arrangements of 30 people in 30 different positions = 30!.
The arrangements in which we have fixed first, second, third & fourth positions (that is, Mark is first, Colleen is Second, Jeff is third & Emily is fourth) = All possible arrangements of 26 people( all people except Mark, Colleen, Jeff & Emily) in 26 different(from 5th to 30th position) = 26!.
So required probability = 26!/30! = 1/(27 x 28 x 29 x 30).
b) In this case, winning the first prize, winning the second prize etc are independent events as for any person winning a prize do not affect the chances of winning another prize.
So P(Mark is first) = 1/30.
P(Colleen is Second) = 1/30.
P(Jeff is third) = 1/30.
P(Emily is fourth) = 1/30.
So the required probability = P(Mark is first) AND P(Colleen is Second) AND P(Jeff is third) AND P(Emily is fourth)
= (1/30)x(1/

30)x(1/30)x(1/30) = 1/(30^4).

2. Answer :

I will be preffering two dice instead of three, as probability of getting 8 using two dice is higher than that of getting 8 using three dice.
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Two Dice, Probability of getting 8:
Out of 6x6 possibilities, 5 possibilities will result in a sum of eight {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)}.
So Probability = 5/36.
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Three Dice, Probability of getting 8:
Out of 6x6x6 possibilities, 21(= 3 + 6 + 6 + 3 + 3) will result in a sum of eight using different permutations of elements of the set {(1, 1, 6), (2, 1, 5), (1, 3, 4), (2, 2, 4), (3, 3, 2)}.
How 21??
We have to produce 8 using three dice with a constraint that each die must produce a number greater than or equal to 1.
So let's give 1, 1, 1 to all each of the three dice.
Now we are left with 5(i.e. 8 - (1 + 1 + 1)), we need to divide this sum of 5 into 3 dice.In how many ways this can be done?
This problem is similar to counting the number of ways of distributing 5 chocolates among 3 children where each child can have 0 or more chocolates.
I am using * to denote chocolates, and | as separators(we need two separators for 3 children).
for example **|*|** means First child got 2 chocolates, second got 1 and the third one got 2 chocolates.
*|****| means first child got 1 chocolate, second got 4 and the third one got 0 chocolates.
||***** means first child got 0 chocolate, second got 0 and the third one got 5 chocolates.
So the ways to arrange these 7 items(that is 5 chocolates and 2 separators) is 7!/(5!x2!) =21.
So Probability = 21/216.
___________________________
As (5/36) > (21/216), hence we can expect 8 with a higher probability using two dice.

3 (i) Answer :

3(i)
Sample space: All the bit strings of length 10 = 2^10 = 1024.
a) P(0) = P(1) = 0.5
So required probability = (P(1))^10 = (0.5)^10 = 1/1024.
b) P(1) = 0.7
So required probability = (P(1))^10 = (0.7)^10.
c) In part a) and b), the probability of getting 1 at any place in the bit string was indpendent of the position of that place in the bit string.But here it is dependent.
So Probability of getting a 1 at first bit position = 1/(2^1),
Probability of getting a 1 at second bit position = 1/(2^2),
Probability of getting a 1 at third bit position = 1/(2^3),
and so on...
So required probability = (1/(2^1))x(1/

(2^2))x(1/(2^3))x...x(1/(2^10)) = (1/2^(1 + 2 + 3 + ... + 10)) = (1/2^55).
Similarly you can solve 3(ii).

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