Question

For an urn containing 4 red balls and 6 green balls, let the number of balls randomly drawn be the number of heads turning up when 5 fair coins have been previously flipped. What is the probability of drawing 3 green balls?

why do we need binomial theorem for this

Answer #1

(we require binomial distribution for number of heads which on 5 flips since number of trails n=5 are fixed and probability of head is constant and independent from trail to trail)

probability of drawing 3 green balls

=P(3 heads)*P(3 green|3 heads)+P(4 heads)*P(3 green|4 heads)+P(5 heads)*P(3 green|5 heads)

=(_{5}C_{3})*(0.5)^{3}*(0.2)^{2}*((_{6}C_{3})*(_{4}C_{0})/(_{10}C_{3}))+(_{5}C_{4})*(0.5)^{4}*(0.2)^{1}*((_{6}C_{3})*(_{4}C_{1})/(_{10}C_{4}))+(_{5}C_{5})*(0.5)^{5}*(0.2)^{0}*((_{6}C_{3})*(_{4}C_{2})/(_{10}C_{5}))

=0.3125*0.1667+0.15625*0.3810+0.03125*0.4762

**=0.1265**

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List the possible values for X and then find the entire probability
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A. Let X = the number of green balls in Urn A after this process.
List the possible values for X and then find the entire probability
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A. Let X = the number of green balls in Urn A after this process.
List the possible values for X and then find the entire probability
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A. Let X = the number of green balls in Urn A after this process.
List the possible values for X and then find the entire probability
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From an urn containing 9 red balls and 6 green balls,
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