i)Floyd-warshall: Floyd-Warshall Algorithm is an algorithm for finding the shortest path between all the pairs of vertices in a weighted graph. This algorithm works for both the directed and undirected weighted graphs. But, it does not work for the graphs with negative cycles (where the sum of the eRead more
i)Floyd-warshall:
Floyd-Warshall Algorithm is an algorithm for finding the shortest path between all the pairs of vertices in a weighted graph. This algorithm works for both the directed and undirected weighted graphs. But, it does not work for the graphs with negative cycles (where the sum of the edges in a cycle is negative).
ii)Bellman-Ford:
The bellman-Ford algorithm helps us find the shortest path from a vertex to all other vertices of a weighted graph.
Bellman-Ford algorithm works by overestimating the length of the path from the starting vertex to all other vertices. Then it iteratively relaxes those estimates by finding new paths that are shorter than the previously overestimated paths.
Four coins tossed so no of outcome will be 16 Let x be the no of heads so, x can take value 0 to 4 as the minimum head we get is 0 and max 4 P(x=0)= nCr = 4C0 = 1/total outcome = 1/16 same, P(x=1) = 4/16 P(x=2) =6/16 P(x=3) = 4/16 P(x=4) =1/16
Four coins tossed so no of outcome will be 16
Let x be the no of heads
so, x can take value 0 to 4 as the minimum head we get is 0 and max 4
P(head for one coin) = 1/2 so for 6 times head P(head) = (1/2)^6 = 1/64 mean = np = 6400 x 1/64 = 100 (where n=6400 and p = 1/64) Now, poisson distribution formula is = (e-m) (mx) / x! so, poisson distribution of 6 head x times = (e-100) (100x) / x!
P(head for one coin) = 1/2
so for 6 times head P(head) = (1/2)^6 = 1/64
mean = np = 6400 x 1/64 = 100 (where n=6400 and p = 1/64)
Now, poisson distribution formula is = (e-m) (mx) / x!
so, poisson distribution of 6 head x times = (e-100) (100x) / x!
Min-max is a decision-making algorithm that uses decision theory, game theory, statistics, and philosophy to calculate the optimal move It is a two-player game. The mechanism evaluates minimum loss and maximum profit. This logic can also be extended to play the more complicated games like chess, cheRead more
Min-max is a decision-making algorithm that uses decision theory, game theory, statistics, and philosophy to calculate the optimal move It is a two-player game. The mechanism evaluates minimum loss and maximum profit. This logic can also be extended to play the more complicated games like chess, checkers, etc.
In the Tic-Tac-Toe game, a player tries to ensure two cases:
• Maximize a player’s own chances of winning.
• Minimize the opponent’s chances of winning.
Maximize profit: The profit can be maximized by either fork or win.
Fork: Initially player will create an opportunity where he can win in two ways.
Win: If there are two same X or O in a row, then play the third to get three in a row.
Minimize Loss: The loss can be minimized by a block.
Block: If two ‘x’ or ‘o’ of the opponent are in a row then block it, or else block the opponent’s fork.
Note: Download the above pic by clicking on the attachment button below
Q//1. (i) not more than three will be busy? answer: (i) not more than three will be busy? answer: Solve for P(x<=3) = P(x=0) + P(x=1)+ P(x=2)+P(x=3) using formula nCr * (p)^r * (q)^n-r where n= 6 , p = 1/15 and q= 14/15 P(x<=3) = [1-P(x≥3)]+P(x=3) P(x<=3) =0.99492+[6C3 * (1/15)^3 * (14/15)Read more
Q//1. (i) not more than three will be busy?
answer:
(i) not more than three will be busy?
answer:
Solve for P(x<=3) = P(x=0) + P(x=1)+ P(x=2)+P(x=3)
Six coins are tossed 6400 times. Using the poisson distribution, determine the approximate probability of getting six heads x times.
https://sikshapath.in/question/six-coins-are-tossed-6400-times-using-the-poisson-distribution-determine-the-approximate-probability-of-getting-six-heads-x-times-2/
Illustrate working of the following with an example: i) Floyd Warshall Algorithm ii) Bellman Ford Algorithm
i)Floyd-warshall: Floyd-Warshall Algorithm is an algorithm for finding the shortest path between all the pairs of vertices in a weighted graph. This algorithm works for both the directed and undirected weighted graphs. But, it does not work for the graphs with negative cycles (where the sum of the eRead more
i)Floyd-warshall:
Floyd-Warshall Algorithm is an algorithm for finding the shortest path between all the pairs of vertices in a weighted graph. This algorithm works for both the directed and undirected weighted graphs. But, it does not work for the graphs with negative cycles (where the sum of the edges in a cycle is negative).
ii)Bellman-Ford:
The bellman-Ford algorithm helps us find the shortest path from a vertex to all other vertices of a weighted graph.
Bellman-Ford algorithm works by overestimating the length of the path from the starting vertex to all other vertices. Then it iteratively relaxes those estimates by finding new paths that are shorter than the previously overestimated paths.
See lessIn FOUR tossed of a coin , let x be the number of heads. calculate the expected values of x.
Four coins tossed so no of outcome will be 16 Let x be the no of heads so, x can take value 0 to 4 as the minimum head we get is 0 and max 4 P(x=0)= nCr = 4C0 = 1/total outcome = 1/16 same, P(x=1) = 4/16 P(x=2) =6/16 P(x=3) = 4/16 P(x=4) =1/16
Four coins tossed so no of outcome will be 16
Let x be the no of heads
so, x can take value 0 to 4 as the minimum head we get is 0 and max 4
P(x=0)= nCr = 4C0 = 1/total outcome = 1/16
same,
P(x=1) = 4/16
P(x=2) =6/16
P(x=3) = 4/16
P(x=4) =1/16
See lessSix coins are tossed 6400 times. Using the poisson distribution, determine the approximate probability of getting six heads x times.
P(head for one coin) = 1/2 so for 6 times head P(head) = (1/2)^6 = 1/64 mean = np = 6400 x 1/64 = 100 (where n=6400 and p = 1/64) Now, poisson distribution formula is = (e-m) (mx) / x! so, poisson distribution of 6 head x times = (e-100) (100x) / x!
P(head for one coin) = 1/2
so for 6 times head P(head) = (1/2)^6 = 1/64
mean = np = 6400 x 1/64 = 100 (where n=6400 and p = 1/64)
Now, poisson distribution formula is = (e-m) (mx) / x!
so, poisson distribution of 6 head x times = (e-100) (100x) / x!
See lessShow the working of the Minimax algorithm using Tic-Tac-Toe Game.
Min-max is a decision-making algorithm that uses decision theory, game theory, statistics, and philosophy to calculate the optimal move It is a two-player game. The mechanism evaluates minimum loss and maximum profit. This logic can also be extended to play the more complicated games like chess, cheRead more
Min-max is a decision-making algorithm that uses decision theory, game theory, statistics, and philosophy to calculate the optimal move It is a two-player game. The mechanism evaluates minimum loss and maximum profit. This logic can also be extended to play the more complicated games like chess, checkers, etc.
In the Tic-Tac-Toe game, a player tries to ensure two cases:
• Maximize a player’s own chances of winning.
• Minimize the opponent’s chances of winning.
Maximize profit: The profit can be maximized by either fork or win.
Fork: Initially player will create an opportunity where he can win in two ways.
Win: If there are two same X or O in a row, then play the third to get three in a row.
Minimize Loss: The loss can be minimized by a block.
Block: If two ‘x’ or ‘o’ of the opponent are in a row then block it, or else block the opponent’s fork.
Note: Download the above pic by clicking on the attachment button below
See lessQuestion 1 2 Points Assume that on an average one telephone number out of fifteen is busy. What is the …
Q//1. (i) not more than three will be busy? answer: (i) not more than three will be busy? answer: Solve for P(x<=3) = P(x=0) + P(x=1)+ P(x=2)+P(x=3) using formula nCr * (p)^r * (q)^n-r where n= 6 , p = 1/15 and q= 14/15 P(x<=3) = [1-P(x≥3)]+P(x=3) P(x<=3) =0.99492+[6C3 * (1/15)^3 * (14/15)Read more
Q//1. (i) not more than three will be busy?
answer:
(i) not more than three will be busy?
answer:
Solve for P(x<=3) = P(x=0) + P(x=1)+ P(x=2)+P(x=3)
using formula nCr * (p)^r * (q)^n-r
where n= 6 , p = 1/15 and q= 14/15
P(x<=3) = [1-P(x≥3)]+P(x=3)
P(x<=3) =0.99492+[6C3 * (1/15)^3 * (14/15)^3]
P(x<=3) =0.99492+0.00481=0.99973
(ii) at least three of them will be busy ?
Answer: P(x≥3) = P(x=3) + P(x=4)+ P(x=5)+P(x=6)
P(x≥3)= P(x=3) + P(x=4)+ P(x=5)+P(x=6)
P(x≥3)=[6C3 * (1/15)^3 * (14/15)^3 ]+[6C4 * (1/15)^4 * (14/15)^2 ]+[6C5 * (1/15)^5 * (14/15)^1 ]+[6C6 * (1/15)^6 * (14/15)^0 ]
P(x≥3)=0.00508
Q2// answer: go through the below link
https://sikshapath.in/question/out-of-800-families-with-4-children-each-how-many-families-would-be-expected-to-have-1-two-boys-2/
Q//3. answer:
https://sikshapath.in/question/the-heights-of-persons-with-no-of-persons-is-given-heights-in-cm-58-59-60-61-62/
Q//4. answer: click on sample case to get pdf —>sample case
Q//5. answer: click on bolt factory to get pdf —> bolt factory
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