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In $1$ bag there are $22$ red balls and $19$ green balls. Answer this.

If $12$ balls are drawn from the bag, determine the probability that at least $8$ drawn out are red? You are to assume that each ball is returned to the bag after each draw. Show appropriate calculation to support your answer.

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closed as off-topic by user296602, Shailesh, Claude Leibovici, choco_addicted, Edward Jiang May 8 '16 at 4:16

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    $\begingroup$ this is a homework problem. what have you tried ? $\endgroup$ – amanuel2 May 8 '16 at 2:10
  • $\begingroup$ No, it is for an assignment, problem solving question $\endgroup$ – Veljko May 8 '16 at 2:11
  • $\begingroup$ I'm just really stuck and don't know how to work out using calculations $\endgroup$ – Veljko May 8 '16 at 2:15
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    $\begingroup$ The two (assignment & homework) are considered the same. Given it's an item of assessment, it's a good idea to show some effort and research! $\endgroup$ – Rubicon May 8 '16 at 2:15
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    $\begingroup$ Can you compute the probability that exactly $8$ red ones are drawn? $\endgroup$ – lulu May 8 '16 at 2:16
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First of all, you need to choose the order that you draw the x red balls.

You have ${12\choose x}$ different orders to draw x red balls from the bag.

In each draw, you have a $\frac{22}{41}$ chance to draw a red ball and $\frac{19}{41}$ to draw a blue ball.

So the probability to choose x red balls is ${12\choose x}\times (\frac{22}{41})^{x}\times (\frac{19}{41})^{12-x}$

So, $P$ (at least 8 red balls) = $\sum_{i=8}^{12}{12\choose i}\times (\frac{22}{41})^{i}\times (\frac{19}{41})^{12-i}$

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  • $\begingroup$ This would appear to assume that the balls are drawn without replacement. $\endgroup$ – lulu May 8 '16 at 2:24
  • $\begingroup$ @lulu you are right. i'll fix that. thanks $\endgroup$ – Liam May 8 '16 at 2:25
  • $\begingroup$ @lulu just fixed it $\endgroup$ – Liam May 8 '16 at 2:33
  • $\begingroup$ Looks good (+1). $\endgroup$ – lulu May 8 '16 at 2:36

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