Four draws are made with replacement from a bowl that contai

Four draws are made with replacement from a bowl that contains 10 candies: 7 red and 3 green. (With replacement means the candies are drawn one at a time, and each drawn candy is replaced and the candies remixed before the next draw).

a) What\'s the probability all four candies will be red?

b) Find the mean u and the standard deviation o for the number of red candies drawn.

Please show how you arrived at answer, I\'m having trouble with the process. I assigned extra points since there are two parts to the question. Thanks :)

Solution

Probability of drawing a red candy = 7/10

Probability all four candies drawn will be red = (7/10) * (7/10) * (7/10) * (7/10)

= 0.2401

(Since each draw is independent of the other draw)

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Now, let X be the number of red candies out of the 4 draws.

Now, X follows Binomial distribution with n = 4 and p = 0.7

P(X = 0) = (4 C 0)*0.7⁰*(1-0.7)⁴ = (3/10) * (3/10) * (3/10) * (3/10) = 0.0081

P(X = 1) = (4 C 1)*0.7¹*(1-0.7)³ = 0.0756

P(X = 2) = (4 C 2)*0.7²*(1-0.7)² = 0.2646

P(X = 3) = (4 C 3)*0.7³*(1-0.7)¹ = 0.4116

P(X = 4) = (4 C 4)*0.7⁴*(1-0.7)⁰ = 0.2401

Mean number of red candies = summation (X * P(X) )

= 0*0.0081 + 1*0.0756 + 2*0.2646 + 3*0.4116 + 4*0.2401

= 2.8

Variance = summation ((X - mean)² * P(X) )

= (0 - 2.8)² * 0.0081 + (1 - 2.8)² * 0.0756 + (2 - 2.8)² * 0.2646 + (3 - 2.8)²*0.4116 + (4 - 2.8)² * 0.2401

= 0.84

Standard deviation = sqrt(0.84)

= 0.9165

Hope this helped. I will feel happy if you assign extra points :)