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3 years ago

11

Probability check point 6 , you have a bag containing 2 red balls, 4 green balls, and 5 purple balls. you draw one ball from the

bag: it's green. call this draw 0. keep this ball and continue drawing (without replacement) until you get another green ball. call these draws 1, 2...k. what is the probability that you don't draw another green ball until draw 3

1 answer:

natta225 [31]3 years ago

8 0

There are now 10 balls in the bag since you took 1 green out
There are now 3 greens in the bag
3/10 is the chance you pick a green,so 7/10 is the chance you DIDNT pick a green ball for the first round
Second round:-1 ball that is not green now the number of ball is 9
10+9=19

(# of green balls still in the bag)3/19 final answer

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Part a)

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We apply the continuity correction factor to get;

P(X<24+0.5)=P(X<24.5)

We convert to z-scores to get:

$P(X \: < \: 24.5) = P(z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P(z \: < \: 1.20) \\ = 0.8849$

Part c)

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Apply correction factor to get:

P(X>11-0.5)=P(X>10.5)

We convert to z-scores:

$P(X \: > \: 10.5) = P(z \: > \: \frac{10.5 - 19.29}{4.33} ) \\ = P(z \: > \: - 2.03)$

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We want to find the probability that:

between 24 and 28, inclusive, will gain weight as a side effect.

P(24≤X≤28)=

P(23.5≤X≤28.5)

Convert to z-scores:

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