To find the probability that an event will not occur, you work out the probability that it will occur, and then take this number away from 1. For example, the probability of not rolling two 6s in a row can be worked out the following way:The probability of rolling two 6s in a row is 1/6 x 1/6 = 1/36Thus the probability of not rolling two 6s in a row is 1 - 1/36=35/36.
The number of 6s in 37 rolls of a loaded die and binomial.
! So first of all we need to calculate the probability of the base event, that is to get a pair when rolling 2 dice. This is quite simple and we can see it in 2 ways. a) We throw the first dice, regardless of what it comes up as, we have 1 in 6 chanses that the second dice matches it i.e probability = 1/6 b) There are 6*6 possible out comes when rolling two dice. Out of those we can get pair of 1s, pair of 2s , pair of 3s, pair of 4s, pair of 5s or pair of 6s. That is 6 pairs out of 36 total = 1/6 probability. Now about making a sequence of throws. The probability of us making the pair in the first throw is as the basic event = 1/6. If we make it we dont continue and have reached our goal! If we dont get a pair (proability 1-1/6 = 5/6) we continue and make our second throw. Again we have a 1/6 chanse, so in total that we make it in exactly the 2nd throws is 5/6 (miss)*1/6 (hit) = 5/36. If we dont make it , we continue again. Now our chance in making it in exactly the 3rd time is 5/6 (miss)*5/6 (miss again) *1/6 (hit) = 25/216. So the probability of making a pair in three or fewer rolls is the sum of the above, so 1/6 + 5/36 + 25/216 = 91/216 = 0.42 (roughly) Another way when you have established the probability of a single action to find how many repeats you need is to use the Binomial or Poisson distribution (look it up). Best Regards
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
There are four 6's in a standard deck of 52 cards.
1 in 36
To find the probability that an event will not occur, you work out the probability that it will occur, and then take this number away from 1. For example, the probability of not rolling two 6s in a row can be worked out the following way:The probability of rolling two 6s in a row is 1/6 x 1/6 = 1/36Thus the probability of not rolling two 6s in a row is 1 - 1/36=35/36.
Zero. If you roll five dice, you cannot get six 6s.
The answer, using the Binomial distribution is 100C5*(1/6)5*(5/6)95 = 0.000291 (approx).
0
The number of 6s in 37 rolls of a loaded die and binomial.
! So first of all we need to calculate the probability of the base event, that is to get a pair when rolling 2 dice. This is quite simple and we can see it in 2 ways. a) We throw the first dice, regardless of what it comes up as, we have 1 in 6 chanses that the second dice matches it i.e probability = 1/6 b) There are 6*6 possible out comes when rolling two dice. Out of those we can get pair of 1s, pair of 2s , pair of 3s, pair of 4s, pair of 5s or pair of 6s. That is 6 pairs out of 36 total = 1/6 probability. Now about making a sequence of throws. The probability of us making the pair in the first throw is as the basic event = 1/6. If we make it we dont continue and have reached our goal! If we dont get a pair (proability 1-1/6 = 5/6) we continue and make our second throw. Again we have a 1/6 chanse, so in total that we make it in exactly the 2nd throws is 5/6 (miss)*1/6 (hit) = 5/36. If we dont make it , we continue again. Now our chance in making it in exactly the 3rd time is 5/6 (miss)*5/6 (miss again) *1/6 (hit) = 25/216. So the probability of making a pair in three or fewer rolls is the sum of the above, so 1/6 + 5/36 + 25/216 = 91/216 = 0.42 (roughly) Another way when you have established the probability of a single action to find how many repeats you need is to use the Binomial or Poisson distribution (look it up). Best Regards
The chance of throwing a six with one die is one in six. With a pair of dice it is one in six2 or one in 36.
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
-40 = 8-6s -40 - 8 = 6s -48 = 6s 8 = s
This question is tricky because distance and location are not well defined within the electron cloud. So for the sake of explanation, let us assume that the question is "Which has a region maximum probability closest to the nucleus?" I could not find an answer in print for the 4f vs 6s. However, I did find an answer comparing the 3d and 4s. I believe that an understanding of a comparison between the 3d and 4s can be extended to a comparison between the 4f and 6s. The most probable distance of the 3d is less than the 4s. Therefore, we could extend this and say that most likely the 4f sublevel would have a distance of maximum probability closer than the 6s. You did not ask why so I will not go into an explanation.
-6s+10 = -5s 5s-6s = -10 -s = -10 s = 10