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probability

Probabilities and exercises

I've been practicing some probabilities for last few weeks around 10/week usually did the job, however as the topics got harder I've started struggling and now I'm completely stuck. I've been looking online and i did find similar examples but nothing to touch my case in particular. I will continue looking for an answer so even if u can't answer my specific cases links to online literature will be appreciated.

answers are welcome, however i would rather be more interested in explanation of how a problem works.

  1. urn contains m white and k black balls. two players are drawing the balls one after another without putting them back into the urn. the winner is the first person to draw white ball. what is the probability that second player will be the winner? (k=4, m=4)
  2. women tend to vote with probability of a, men tend to do the same with probability of b. the probability c tells us that if we take a couple one of them will not go voting. what are the chances that at least one of them will vote? (a=0.49, b=0.61, c=0.75)
  3. we are sending a message of n bytes. to obtain higher chance of sending entire message without ruining it we use k different wires. What is the probability of sending an entire message through one of the wires without ruining it, if the probability of ruining any of the bytes at any of the wires is p. (p=0.06, n=7, k=6)
  4. the basketball game finals play to N wins. after m + n games the result is m : n. what is the probability of winning the finals for the lagging(the team that's behind) team if it is known that the leading team wins each match with a probability of p? (m=3 n=2 N=5 p=0.36)

sorry for my English and any help will be appreciated

Solution

  • Just A

    So Given m = k and both = 4. And each player takes a ball out per turn. Given the worst case scenario 4 good balls will be picked (K) there will be 2 turns of both players picking the black ball then the first player will be guaranteed the white ball next turn. Therefore you need to work out the probability for the first two turns.

    Turn 1 - P1 turn 1 Chance for black= 4/8 - P2 turn 2 Change for white = 4/7 Therefore turn 1 chance = (4/8)*(4/7). = 28.57%

    Turn 2 - P1 turn 1 Chance for black= 2/6 - P2 turn 2 Change for white = 4/5 Therefore turn 2 chance = (4/6)*(4/5). = 26.664% but to get to turn 2 is a 28.57% chance

    Therefore its the first probability add the second probability given the first case occurs. So 28.57% + (26.44%*28.57%)= 36.12%

    The third case is a always lose. Top tip then be the first player! Turn 3 Instance win for player 2 therefore a loss. = 0%