Find the probability of choosing exactly 4 red cards

[chwala]

Homework Statement

see attached.

Relevant Equations

hypergeometric distribution

Find the question and solution here ( sorry its a bit blurred) ... given using the hypergeometric method...i wanted to understand what is the clear distinction between hypergeometric and binomial distribution? Could it be in in reference to no replacement and replacement...?

My approach on this question;
Let $R$ = red and $B$= black.
$P_r(RRRRB)$= $\dfrac {6}{20}$×$\dfrac {5}{19}$×$\dfrac {4}{18}$×$\dfrac {3}{17}$×$[5×$$\dfrac {14}{16}]$=$0.0135$ as required. Black can be arranged in $5$ different ways...cheers

 

[PeroK]

i wanted to understand what is the clear distinction between hypergeometric and binomial distribution? Could it be in in reference to no replacement and replacement...?

You can look these terms up. Binomial distribution assumes the same probability each time, which in this case means the number of red and black cards do not change, which means the card chosen each time must be replaced.

My approach on this question;
Let $r$ = red and $b$= black.
$P(rrrrb)$= $\frac {6}{20}$×$\frac {5}{19}$×$\frac {4}{18}$×$\frac {3}{17}$×$[5×$$\frac {14}{16}]$=$0.0135$ as required. Black can be arranged in $5$ different ways...cheers

Okay.

 

[chwala]

You can look these terms up. Binomial distribution assumes the same probability each time, which in this case means the number of red and black cards do not change, which means the card chosen each time must be replaced.

Okay.

Ok Perok, looks like i have to dedicate time to go through the literature...Math is vast! A whole world by itself and not for the faint of hearts!

 

[chwala]

From my study we can also realize the required result by having,

$P_r[RRRRB]$=$\dfrac{5C4×15C2}{20C6}$=$0.0135$ i.e by using symmetry...

 

[chwala]

I was looking at a similar question, i do not want to start a new thread as the questions are more or less the same...

consider this question;

Now i am trying to use the old school way in finding the solution, actually i am trying to use the tree diagram, now from my tree diagram below

I have the following $3$ possibilities;
$P[bbwwrr]$ or $P[wwrrbb]$ or $P[rrbbww]$.
it follows from my calculations that $P[bbwwrr]$=$P[wwrrbb]$=$P[rrbbww]$.
i.e,
$P[bbwwrr]$=$\dfrac{5×4×10×9×15×14}{30×29×28×27×26×25}$=$\dfrac {1}{1131}$

Now from my reasoning, $bb, ww$ and $rr$ may be chosen each in $2$ ways thus giving us $6$ways... Secondly, when considering say, $P[bbwwrr]$, the first term can be chosen in $1$ way only and the rest can be chosen in $5$ ways therefore we shall have,
$P[bbwwrr]$=$\dfrac {1}{1131}$×$5$×$6$=$\dfrac {30}{1131}$
The required probability value is given by adding,
$P[bbwwrr]$ + $P[wwrrbb]$ +$P[rrbbww]$=$\dfrac {90}{1131}$=$\dfrac {30}{377}$=$0.0795755968$

 

[PeroK]

Now i am trying to use the old school way in finding the solution, actually i am trying to use the tree diagram, now from my tree diagram, i have the following possibilities;

$P[bbwwrr]$ or $P[wwrrbb]$ or $P[rrbbww]$,
it follows that $P[bbwwrr]=P[wwrrbb]=P[rrbbww]$ from my calculations...
i.e,
$P[bbwwrr]$=$\dfrac{5×4×10×9×15×14}{30×29×28×27×26×25}$

If you want to do it that way, you need to count all the different ways you can get two of each (in terms of the sequence the balls are drawn). You have $bbrrww, brbrww, brrbww \dots$. So, a lot more than three ways. Moreover, that all of these are equally likely may need some justification.