Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Mitul
Group Title
Probability of A winning the race is 4/5 and probabilty of B winning the race is 5/6. find the probability that neither wins...
 one year ago
 one year ago
Mitul Group Title
Probability of A winning the race is 4/5 and probabilty of B winning the race is 5/6. find the probability that neither wins...
 one year ago
 one year ago

This Question is Closed

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
dw:1344504997861:dw
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
not possible @Yahoo! 14/55/6 is negative...
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
i think the answer will be 0
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
no cannot be....the event has occured....its not an impossible event...
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
u see there are only two chances either A will win or B will
 one year ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
(14/5)(15/6)
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
that can be one case but i nvr said that only A and B are playing.....
 one year ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
wait these are not mutually exclusive events
 one year ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
oh ok gotcha
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhhh this is just one event right?
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
yeah...event that neither wins...
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
yahoo's solution should've been right...
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
also this doesnt make sense... 4/5 + 5/6 = 49/30
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
the total probability should be less or equal to 1
 one year ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
P(both winning) = 4/5 * 5/6 P(both not winning) = 1 4/5 * 5/6
 one year ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
draw a probability tree. Remember as we go along we multiply the values
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhh no @rsadhvika..
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
what you did is probability that NOT both will win
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
\[P(A \; and \;B \;win) = P(A) \times P(B)\] subtract it from 1 I think..
 one year ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
oh im wrong again @lgbasallote :'(
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
@waterineyes u just did what @rsadhvika did...
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
there must be a silly trick to this :/
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Subtract them from 1 each and then add them..
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
we're thinking too complex...we need to open our brains to the possibilities :/
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
i really dont think this is one event
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
could you post the whole question?/
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
yup @lgbasallote there are more than one case to this problem
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
hmm i knew it =_=
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
this is the whole question @lgbasallote
 one year ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
um, tried my solution?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
the probabilities dont add up to less than or equal to 1
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
i got the answer....
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
i think what @rsadhvika first posted should work then...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
\[P(nor\; A) = \frac{1}{5}\] \[P(nor \; B) = \frac{1}{6}\] \[P(Neither \; A \; nor \;B) = P(nor \; A) + P(nor \; B)\]
 one year ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
THINK SIMPLE! PROBABILITY TREE!
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
Case I Probability A wont win is 1  4/5 = 1/5 Case II probability B wont win is 1  5/6 = 1/6 so the probability both wont win is 1/6 x 1/5 = 1/30
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhh should it be +
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Sorry there will come multiplication..
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
this is not the question: Solution: the first case is when only A and B are playing the second case is that there are other players
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhh can you post the WHOLE question with all these cases?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
no paraphrasing...the exact one...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
I think we are making it complicated.. There is one and one case only..
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
@lgbasallote my qestion is perfect....the above is the solution which i gave for two cases....we need to consider both the probabilities
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
you're suddenly giving situations...so i think what you originally posted isnt the full question
 one year ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
Case 1 1P(A)P(B) Case2...possibility that there are other playersss p(A and B cannot winwin)=1[4/5+5/64/5*5/6]
 one year ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
4/5 > A will win (B cannot win) 1/5 > A will not win (B still can) 5/6 > B will win 1/6 > B will not win (Neither A or B wins) [What you want] If you take a look at my probability tree, you will see what I mean. Remember you multiply the values as you go along. So, YOUR ANSWER > 1/6 * 1/5 = 1/30
 one year ago

telliott99 Group TitleBest ResponseYou've already chosen the best response.0
The example violates basic probability rules. If only one can win then P(A) + P(B) <= 1
 one year ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
@telliott99, remember there are others in this 'race' or whatever it is.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.