ride the bus probability

With probability $\frac{4}{52}$ it transitions into state $(10,1,0,0,0)$ which involves a drink.

$$P_2 = 2\sum_{i=7}^{12}\frac{i}{13}\cdot\frac{1}{13} + \frac{6}{13}\cdot\frac{1}{13}$$$$P_2 = \frac{2}{169}\sum_{i=7}^{12}i + \frac{6}{169}$$$C_n = \sum_{i=1}^{13-n}(P[\textrm{Draw card }(i+n)] \cdot P[\textrm{Correct guess given the difference}])$$C_n = \sum_{i=1}^{6}\frac{1}{13-n} \cdot \frac{12-i}{13} + \sum_{i=7}^{13-n}\frac{1}{13-n} \cdot \frac{i-1}{13}$$C_n = \frac{1}{13(13-n)}(\sum_{i=1}^{6}(12-i) + \sum_{i=7}^{13-n}(i-1))$$C_n = \frac{1}{13(13-n)}(12(6) - \sum_{i=1}^{6}(i) + \sum_{i=7}^{13-n}(i) - (13-n-7+1))$$C_n = \frac{1}{13(13-n)}(72 - 21 + \sum_{i=1}^{7-n}(i+6) - (7-n))$$C_n = \frac{1}{13(13-n)}(51 + \sum_{i=1}^{7-n}(i) + 6(7-n) - (7-n))$$C_n = \frac{1}{13(13-n)}(51 + \frac{(7-n)(8-n)}{2} + 5(7-n))$$C_n = \frac{1}{13(13-n)}(51 + \frac{n^2 - 15n + 56}{2} + 35-5n)$$C_n = \frac{1}{13(13-n)}(86 + \frac{n^2 - 15n + 56}{2} - 5n)$$C_n = \frac{1}{26(13-n)}(172 + n^2 - 15n + 56 - 10n)$$C_7 = \frac{1}{13(13-7)}(\sum_{i=1}^{6}(12-i) + \sum_{i=7}^{13-7}(i-1))$$P_3 = \sum_{i=1}^{13} C_i P[\textrm{First card} = i]$$P_3 = \frac{1}{13}(\frac{17}{26} + 2(\frac{17}{26} + \frac{7}{11} + \frac{81}{130} + \frac{8}{13} + \frac{8}{13} + \frac{57}{91}))$$P_1 P_2 P_3 P_4 = \frac{1}{2} \cdot \frac{120}{169} \cdot \frac{82029}{130130} \cdot \frac{1}{4} = \frac{246087}{4398394} \approx 0.055949 \approx \frac{1}{17.87}$ Learn more about Stack Overflow the company

2. By clicking “Post Your Answer”, you agree to our To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Buses arriving within very small (bus bunching) or very large headways are of particular concern and much less desirable.

Given the condition of the machine, the probability of producing a defective item is known. When you ride the bus you are late 30% of the time and when you ride the train you are late 20% of the time. By using our site, you acknowledge that you have read and understand our Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields.

That is, assume that you have deck of 52 cards and you start knowing that you have to flip $10$.

The transit authority would not have to do anything special to the bus, no additional fuel would be needed, and yet the transit authority could charge triple the price. If multiple players are tied with the same amount of cards, one will be selected randomly.

drinks and two cards. 3.Teri rides the bus to school if she gets up by 7:30 A.M. Teri gets up at 7:30 A.M. about half of the time so can you estimate the probability that Teri will ride the bus to school?

Note: The model does not … With the remaining probability you transition to state $(9,0,0,0,0)$ with no drink.Hence value of $(10,0,0,0,0)$is $$v_{(10,0,0,0,0)}=\tfrac{4}{52}(1+v_{(10,1,0,0,0)})+\tfrac{4}{52}(2+v_{(11,0,1,0,0)})+\tfrac{4}{52}(3+v_{(12,0,0,1,0)})+\tfrac{4}{52}(4+v_{(13,0,0,0,1)})+\tfrac{52-16}{52}v_{(9,0,0,0,0)}$$ If you know the value of the states in the expression, you know the expected number of drinks. Check your Now, what about those outcomes that are not the expected or favorable event. You draw a card. Learn more about Stack Overflow the company
2nd guess: higher or lower than first card 3rd guess: inside, outside, or same as first two 4th guess: color 5th guess: higher or lower (than the 4th card) 6 guess: suit

and put the card back among the $52$.With this, state is described simply by the number of card you have to draw.Thanks for contributing an answer to Mathematics Stack Exchange! It is played by guessing cards.

It is totally underrated if you ask me. 2. Detailed answers to any questions you might have Once you answer all four questions correctly in a row, you're done.Since we're drawing with replacement, you always have a Therefore, the probability of guessing correctly here isOur choice for this stage depends on the difference between the ranks of the two drawn cards. There are several questions like this on my assignment and I have had no luck in figuring out how to work them after several hours of looking through class notes and on the internet. Assign ranks Given the above guessing scheme, here are the probabilities of guessing correctly for each possible difference between the first two ranks:The distribution is naturally mirrored for higher cards, where The maximum difference between the first two cards if the first card is If a player is not counting cards in any way, they should guess as such:Thanks for contributing an answer to Mathematics Stack Exchange!

Decently ventilated public transit mode with minimal talking and movement.

Assume the arrival times of the bus and passenger are independent of one another and that the passenger will wait up to $15$ minutes for the bus to arrive.
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