Simple Probability Question [*RESOLVED*]

[getnandu]

Hi !

I am not good at maths like you experts, that why I am taking Maths classes.
Recently I came across the following two Probability questions and i am not able to figure out how to solve them.

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Q1.) If four coins are tossed, find the probability of
the occurrence of 2 heads and 2 tails ?

Q2.) The odds that a book will be favourably reviewed by
three independent critics are 3 to 2, 4 to 3 and 2 to
3 respectively. What is the probability that of the
three reviews majority will be favourable?
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Any Help will be appreciated.

Thanks
Nandu

[DiGiTaIErRoR]

Originally posted by getnandu

Q1.) If four coins are tossed, find the probability of
the occurrence of 2 heads and 2 tails ?

My program says ~.375.

Dim x As Long, y As Long
Dim flip As Long
Dim h As Long, t As Long, tc As Long
Randomize
 
For y = 1 To 1000000
    For x = 0 To 3
        flip = Int(2 * Rnd)
        If flip = 1 Then h = h + 1 Else t = t + 1
    Next
    If h = 2 And t = 2 Then tc = tc + 1
    h = 0
    t = 0
Next
MsgBox tc / 1000000

But my math says 15/16. 1 - (1/2)^4

[mendhak]

Originally posted by getnandu


Q2.) The odds that a book will be favourably reviewed by
three independent critics are 3 to 2, 4 to 3 and 2 to
3 respectively. What is the probability that of the
three reviews majority will be favourable?
-----------------------------------------------------------------

3/5 * 4/7 * 2/5

= 24/175

= 0.137

[ZaidGS]

make a tree !!!

add starred values (**)
(each has value 0.5^4 )

** head
**** head
****** head
******** head
******** tail
****** tail
******** head
******** tail (**)
**** tail
****** head
******** head
******** tail (**)
****** tail
******** head (**)
******** tail
** tail
**** head
****** head
******** head
******** tail (**)
****** tail
******** head (**)
******** tail
**** tail
****** head
******** head (**)
******** tail
****** tail
******** head
******** tail
so the probability is ( 0.5 ^ 4 * 6 )
=37.5%

[kedaman]

Q1:

2^4 equally possible outcomes, out of which combinations of two in four is 4c2
4c2/2^4=3/8

Q2:

2 or 3 makes a majority out of 3 and one does for a minority, so if we assume that the minority is favourable and subtract its probability from 1 we have the probability for the majority. Each of the critics could be the favouring critic while the others are not, thus:
1 - (3/5*3/7*3/5 + 2/5*4/7*3/5 + 2/5*3/7*2/5)=0.64

[kedaman]

erm.. forgot that if none is in favour then it is also a minority

1 - (3/5*3/7*3/5 + 2/5*4/7*3/5 + 2/5*3/7*2/5+2/5*3/7*3/5)=0.537143

[jemidiah]

Explain what? It comes out approximately correct (the 0.537143 is for the second question)

[getnandu]

Thanks very much guys for your help.
You guys are right.

Ans 1.) 3/8
Ans 2.) 94/175

[getnandu]

How do I change my initial thread subject to include (resolved)

I did not find any option to do that !!!


Nandu

[kedaman]

click on edit to the bottom right of your first post and change the subject