Question

Probability
attached image

Answer 1

@rational
@thomas5267

Answer 2

\(5\times\dfrac{1}{2^3}\)? The probability of winning three times is \(\dfrac{1}{2^3}\) and there are 5 position for this three consecutive win to occur.

IIIXXXX
XIIIXXX
XXIIIXXX
XXXIIIX
XXXXIII

Answer 3

Did you get the previous questions right?

Answer 4

sir it was a paragraph of 3 question and it was the second part and idk if it is affected by the first one or not

Answer 5

and it is 'AT LEAST'

Answer 6

and if we are considering only 3 wins and rest all matched should be lost by India.. so the denom will be 2^7
so basically i have the idea of adding 3 wins, 4 wins but my doubt is if the match goes like WWWLWWW

Answer 7

I don't know how to deal with this. Clearly the crux of matter is deriving a formula for 7 objects derangement with n of W and n-7 L with the constraint that 3 W's must be together.

Consider the case n=5 and you will understand the difficulty.
L(WWW)WWL
LW(WWW)WL
LWW(WWW)L
These three configurations are identical and I don't know how to deal with them. Maybe some recursive formula will help?

Answer 8

For five W's:
\[
\underbrace{3!\div 2!}_\text{5 consecutive wins}+\underbrace{2!}_{\substack{\text{4 consecutive wins}\\\text{sandwiched by two loses}}}+\underbrace{2\times2}_{\substack{\text{4 consecutive wins}\\\text{bounded by L}\\\text{and start/end of matches}}}\\+\underbrace{3!\div2!}_{\substack{\text{3 consecutive wins}\\\text{with two side sandwiched by L}}}+\underbrace{2\times3!\div2!}_{\substack{\text{3 consecutive wins}\\\text{bounded by L}\\\text{and start/end of matches}}}\\
=18
\]
The result is computer verified.
I am tired. I have to go to sleep.

Answer 9

what if we try to do it backwards like 1- (no consecutive match won + 2 consecutive matches won) for no consecutive matches won there will not be many cases i guess like 4 W 3L , 3 W 4L, 2W 5L and for 2 consecutive matches I am in doubt

Answer 10

www oooo 5 ways to position 3 wins

5*(1/2)^3*(1/2)^4

wwww ooo 4 ways to position 4 wins

4*(1/2)^4*(1/2)^3

wwwww oo 3 ways to position 5 wins

3*(1/2)^5*(1/2)^2


etc ....

where is my error if it exists?

Answer 11

i have an error of course

Answer 12

www ooow is also a 3 consecutive win setup

Answer 13

we can say that it is at least 15/2^7

can we rule out 47/2^7 ?

Answer 14

still a solution would be good for understanding

Answer 15

none is also an option :P

Answer 16

yeah, i mean we can work it brutely ... i cant come up with anything for calculating the "other" wins that are not consecutive yet ...



www ooow
www oowo
www owoo
www wooo *<-- ignore or exclude of the 4 consecutive win

oww woow
oww wowo
oww wwoo *
www wooo *

oow wwow
oow wwwo *
oww wwoo *
wow wwoo

ooo wwww *
oow wwwo *
owo wwwo
woo wwwo

oow owww
owo owww
woo owww

its a start, you see any patterns from that?

Answer 17

what about 5 wins like wwwowwo

Answer 18

its workable, just a pain lol but yeah, you would want to include the setup for those

Answer 19

there was a bare solution for that
if you understand something, do tell
REQUIRED PROBABILITY 1/2^3+ 1/2^4 + 1/2^4 + 1/2^4 + 1/2^4 -1/2^7

Answer 20

i cant say that i do understand that

Answer 21

w and o each have a 1/2 probability, so all these should add to a denom of 2^7 in my eyes.

Answer 22

lol um completely stuck on this one XD

Answer 23

c = www

assuming we dont want cw or wc

---------------------------
c ooow we have 3 degree of freedom
o c oow we have 2 degrees of freedom
oo c ow we have 2 degrees of freedom
woo c o we have 2 degrees of freedom
wooo c we have 3 degrees of freedom

12 count over all
----------------------------

c ooww we have 3 degrees of freedom
o c oww we have 1 degrees of freedom
wo c ow we have 1 degrees of freedom
wwo c o we have 1 degrees of freedom
wwoo c we have 3 degrees of freedom

9 count over all
-----------------------------

c owww we have 1 degrees of freedom
wwwo c we have 1 degrees of freedom

2 count over all
-----------------------------

22 ways total, did i miss any?

Answer 24

this has not be done by this way.. we were given only 3-4 minutes to solve these kind of questions so making cases doesn't make sense

Answer 25

i dont know of a formula to help out, but i could prolly work it under 5 minutes for this question.

Answer 26

good luck

Answer 27

4 wins:
\[
2!+2\times 2!\div 2!+3!+2\times 3!\div2!=16
\]

Answer 28

3 wins:
\[
5!\div4!=5
\]
7 wins = 1

6 wins:
LWWWLWW
\[
7!\div 6!=7
\]

1+7+18+16+5=47
The probability is \(\dfrac{47}{2^7}\).

Answer 29

For \(3\leq W\leq5\) wins with 3 consecutive wins and 7 matches:\[
\begin{align*}
&\hspace{1.1em} \frac{3!}{(L-2)!(W-3)!}&+2\times\frac{3!}{(L-1)!(W-3)!}\\
&+\frac{2!}{(L-2)!(W-4)!}&+2\times\frac{2!}{(L-1)!(W-4)!}\\
&+\frac{1!}{(L-2)!(W-5)!}&+2\times\frac{1!}{(L-1)!(W-5)!}\\
\end{align*}\\
=\sum_{k=3}^{W}\frac{(6-k)!}{(5-W)!(W-k)!}+2\times\frac{(6-k)!}{(6-W)!(W-k)!}
\]