Probablity question

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

que: On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits (i) A before B? (ii) A before B and B before C? (iii) A first and B last? (iv) A either first or second? (v) A just before B?
probably not
The number of ways by which veena can visit is 4!=24

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yes
i can'r go further
*can't
i guess you can list ways in which a comes before B and count them A bit tedious though
yes that is tediuos i want another time saving way
or if A is the first then how many ways can the others be visited 3*2 = 6 if A is 2nd and B 3rd or 4th - thats 2 ways
- no the last one is 4 ways then we have A 3rd and B 4th - 2 ways total 6 + 4 + 2 = 12 so first one is 12/24 = 1/2
to be honest I hate probability!
sumilarly B before C = 12 ways ?
yes must be
can we multiply these now to get the compound probability?
ok i need to find now the common between A before B and B before C
the question is are they independent events?
i don't understand how independent events means here
there are 12 events in both A before B and B before C and i need to find the common events between them
(A before B) \(\Large \cap\) (B before C)
by listing and counting there are 4 of them
ok so i need to list them necessrily
i'm afraid i dont remember all the set theory applied to probability
ok so same for 3, 4 and 5 , listing ?
questions iii to v are quite straightforward - i think!!
for q iii it is the number ow ways you can arrange the 3 letters between A and B
= 6 right?
3!=6 , yes
for iv A first = 3! ways + 31 ways with A second
* 3!
ok
v AB to begin - - 2 ways AB 2nd and 3rd - 2 waya AB last - 2 ways
nice question
yea it is
AB-- A_B__ A__B _AB__ _A_B_ _A__B __AB i calculate 7 ways for v.)
|dw:1439460306887:dw| just another way of thinking about it"!
is the 7 ways correct for last one ?
No i dont think so - isn't it A just before B (A and B together?)
  • phi
***(i) A before B? *** I guess we could count the different routes but it seems there is nothing special about A and B, and half the time A will be first, and B second (and vice versa)
for (v) A just before B? put A and B next to each other and count it as one item let X=AB then we want the number of arrangements of XCD which is 3!=6 the probability easily follows
  • phi
(ii) A before B and B before C? it seems difficult to find the prob of each event and then subtract off their intersection. Rather, how about writing A B C and asking, where can D go? we can list DABC, ADBC, ABDC, ABCD so 4 patterns out of 24 and prob 4/24= 1/6

Not the answer you are looking for?

Search for more explanations.

Ask your own question