Sanjay rolled a die three times

If three six-sided dies are rolled, once each one, how can I find the probability of getting three even numbers? The number of possible events would be six times six times six? And, if I want to know the probability of any other event, lets say of getting a 6 in any of the three dies, or a addition wich result is a number minor than 14, how can I calculate it? Your first question is basically the same as the one described in the video. Rolling three dice one time each is like rolling one die 3 times. And yes, the number of possible events is six times six times six (216) while the number of favourable outcomes is 3 times 3 times 3. Therefore, the probability is still 1/8 after reducing the fraction, as mentioned in the video. You can calculate the probability of another event just by finding the total number of outcomes, in this case 216, and counting how many results involve getting one or more 6. There are a total of 91 ways to get 6 in at least one of the three dice. Therefore, the probability is 91/216. The same applies for an addition which results in a number less than 14. Count how many results are favourable (produce a number less than 14) and divide that by the total number of outcomes to get the probability. If my calculations are correct, there are 91 ways to get a sum less than 14. The probability will again be 91/216. It can be a hassle to list out and count the individual favourable outcomes. It is important to find some sort of pattern for organized counting...

The sum 15 can be obtained by throwing the dice three times in the following ways, (6,6,3),(6,3,6),(3,6,6) (6,5,4),(6,4,5),(5,4,6),(5,6,4),(4,5,6),(4,6,5) (5,5,5) Therefore, our of total 10 possible ways first roll can be 4 in only 2 ways, Thus, the required probability is 2/10=1/5. Hence, choice(2) is correct. • • : 0 Share the Knowledge You can help us to improve by giving your valuable suggestions at [email protected] By using the service of this site, I agree that I will serve wholeheartedly and will not indulge in any sort of activity that threats the integrity of the organisation I am working for / I work for. I have a firm believe in the notion that knowledge should be open source and helping the needy and deserving part of society will always be my motto.

$\begingroup$ While it does not help at all in figuring out the formula or reasoning behind the result (and that's one of several reasons why I'm not suggesting it as an answer) anydice.com is an excellent tool for either getting immediate result data or exploring the solution space of probability questions like this. Here's a link to your specific question: $\endgroup$ The long winded brute force method would be to add up the probabilities of the four outcomes that would give you the desired result. $$5,5,5\\5,5,x\\5,x,5\\x,5,5\\\text$$ And you would add up these probabilities. The more general way to look at it would be to, look into the binomial distribution, as suggested by a different answer to this question. Also, @robjohn gives a good explanation on how to find the probabilities for three and two 5's. The number of ways to arrange $k$ fives out of $3$ dice is $\displaystyle\binom $$ Since you're performing multiple experiments each with exactly two outcomes (the dice is a five vs. the dice is not a five), this is a classic example of a The standard formula to calculate the probability is then to use (from the $$ P\ $$ As you can see, this method of calculation can unfortunately become very tedious for larger examples, but it has the advantage of being a consistent approach for any problems surrounding Bernoulli trials. The intuitive approaches mentioned in some of the other answers can be quicker for small cases where all the possible outcomes can be manually consid...

• • • • • • • • • About sanjay rolled a die three times Sanjay rolled a die three times and the numbers appearing on the uppermost face when added will be 15 . To find: What is the probability that the first number that appeared was four Solution: Formula used: The formula used to calculate the Probability of an event is given by . People Also Read: If it's a three, then we have 9 ways, . Cotinuing this process will get you the answer — possible ways. How to use If a fair 6 – sided die is rolled three times , what is the probability Total ways in which a 6 sided die can be rolled three times = 6 × 6 × 6 = 2 1 6 to get exactly one 3 , there are three ways ; A 3 on the first roll and non 3 on other two rolls . Sanjay rolled die three times and the numbers appearing on the uppermost face when added will be 15. What is the publ the first number that appeared was four. The probability of getting a larger number than the previous number each time is – This question was previously asked in. RPSC RAS Prelims 27 Oct 2021 Official Paper Download PDF Attempt Online. rix multiplicatio lication find (b) If 10 horses can plough 12, acres of land in 15 days, how many horses will be required to plough 72 acres in 5. Evaluate the expression: Co (180) + 4sin (270) + 3 cos (270) + sec (180) + 4cosec (270): .

That is a result of how he decided to visualize this. Imagine we flip the table around a little and put it into a coordinate system. Along the x-axis you put marks on the numbers 1, 2, 3, 4, 5, 6, and you do the same on the y-axis. We are interested in rolling doubles, i.e. getting the same on both dice. If we let x denote the number of eyes on the first die, and y do the same for the second die, we are interested in the case y = x. But this is the equation of the diagonal line you refer to. At 2.30 Sal started filling in the outcomes of both die. This video wasn't what i was looking for but some of you might be able to help. I had a question: "Two dice are rolled, copy and complete the table below"...... Then there was a table which looked exactly like the one Sal drew. But it had been filled out differently: In the first box (Dice 1 and Dice 2) it had been filled in as 2. Sal wrote 1,1. In another box on my sheet (1 across and three down on Sal's diagram) it had been filled out as 1. I am very confused on how they got this answer so if you understand what I'm talking about please answer. Sorry for the bad explaining! Probably the easiest way to think about this would be: P(Rolling a 1 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 2 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 3 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 4 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 5 four times in a row...

The sum 15 can be obtained by throwing the dice three times in the following ways, (6,6,3),(6,3,6),(3,6,6) (6,5,4),(6,4,5),(5,4,6),(5,6,4),(4,5,6),(4,6,5) (5,5,5) Therefore, our of total 10 possible ways first roll can be 4 in only 2 ways, Thus, the required probability is 2/10=1/5. Hence, choice(2) is correct. • • : 0 Share the Knowledge You can help us to improve by giving your valuable suggestions at [email protected] By using the service of this site, I agree that I will serve wholeheartedly and will not indulge in any sort of activity that threats the integrity of the organisation I am working for / I work for. I have a firm believe in the notion that knowledge should be open source and helping the needy and deserving part of society will always be my motto.

If three six-sided dies are rolled, once each one, how can I find the probability of getting three even numbers? The number of possible events would be six times six times six? And, if I want to know the probability of any other event, lets say of getting a 6 in any of the three dies, or a addition wich result is a number minor than 14, how can I calculate it? Your first question is basically the same as the one described in the video. Rolling three dice one time each is like rolling one die 3 times. And yes, the number of possible events is six times six times six (216) while the number of favourable outcomes is 3 times 3 times 3. Therefore, the probability is still 1/8 after reducing the fraction, as mentioned in the video. You can calculate the probability of another event just by finding the total number of outcomes, in this case 216, and counting how many results involve getting one or more 6. There are a total of 91 ways to get 6 in at least one of the three dice. Therefore, the probability is 91/216. The same applies for an addition which results in a number less than 14. Count how many results are favourable (produce a number less than 14) and divide that by the total number of outcomes to get the probability. If my calculations are correct, there are 91 ways to get a sum less than 14. The probability will again be 91/216. It can be a hassle to list out and count the individual favourable outcomes. It is important to find some sort of pattern for organized counting...

That is a result of how he decided to visualize this. Imagine we flip the table around a little and put it into a coordinate system. Along the x-axis you put marks on the numbers 1, 2, 3, 4, 5, 6, and you do the same on the y-axis. We are interested in rolling doubles, i.e. getting the same on both dice. If we let x denote the number of eyes on the first die, and y do the same for the second die, we are interested in the case y = x. But this is the equation of the diagonal line you refer to. At 2.30 Sal started filling in the outcomes of both die. This video wasn't what i was looking for but some of you might be able to help. I had a question: "Two dice are rolled, copy and complete the table below"...... Then there was a table which looked exactly like the one Sal drew. But it had been filled out differently: In the first box (Dice 1 and Dice 2) it had been filled in as 2. Sal wrote 1,1. In another box on my sheet (1 across and three down on Sal's diagram) it had been filled out as 1. I am very confused on how they got this answer so if you understand what I'm talking about please answer. Sorry for the bad explaining! Probably the easiest way to think about this would be: P(Rolling a 1 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 2 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 3 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 4 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 5 four times in a row...

$\begingroup$ While it does not help at all in figuring out the formula or reasoning behind the result (and that's one of several reasons why I'm not suggesting it as an answer) anydice.com is an excellent tool for either getting immediate result data or exploring the solution space of probability questions like this. Here's a link to your specific question: $\endgroup$ The long winded brute force method would be to add up the probabilities of the four outcomes that would give you the desired result. $$5,5,5\\5,5,x\\5,x,5\\x,5,5\\\text$$ And you would add up these probabilities. The more general way to look at it would be to, look into the binomial distribution, as suggested by a different answer to this question. Also, @robjohn gives a good explanation on how to find the probabilities for three and two 5's. The number of ways to arrange $k$ fives out of $3$ dice is $\displaystyle\binom $$ Since you're performing multiple experiments each with exactly two outcomes (the dice is a five vs. the dice is not a five), this is a classic example of a The standard formula to calculate the probability is then to use (from the $$ P\ $$ As you can see, this method of calculation can unfortunately become very tedious for larger examples, but it has the advantage of being a consistent approach for any problems surrounding Bernoulli trials. The intuitive approaches mentioned in some of the other answers can be quicker for small cases where all the possible outcomes can be manually consid...

• • • • • • • • • About sanjay rolled a die three times Sanjay rolled a die three times and the numbers appearing on the uppermost face when added will be 15 . To find: What is the probability that the first number that appeared was four Solution: Formula used: The formula used to calculate the Probability of an event is given by . People Also Read: If it's a three, then we have 9 ways, . Cotinuing this process will get you the answer — possible ways. How to use If a fair 6 – sided die is rolled three times , what is the probability Total ways in which a 6 sided die can be rolled three times = 6 × 6 × 6 = 2 1 6 to get exactly one 3 , there are three ways ; A 3 on the first roll and non 3 on other two rolls . Sanjay rolled die three times and the numbers appearing on the uppermost face when added will be 15. What is the publ the first number that appeared was four. The probability of getting a larger number than the previous number each time is – This question was previously asked in. RPSC RAS Prelims 27 Oct 2021 Official Paper Download PDF Attempt Online. rix multiplicatio lication find (b) If 10 horses can plough 12, acres of land in 15 days, how many horses will be required to plough 72 acres in 5. Evaluate the expression: Co (180) + 4sin (270) + 3 cos (270) + sec (180) + 4cosec (270): .