Question

. A bag contains 6 red and 3 black chips. One chip is selected, its color is recorded, and it is returned to the bag. This process is repeated until 5 chips have been selected. What is the probability that one red chip was selected?

Answer 1

Answer:

0.0412

Step-by-step explanation:

Total chips = 6 red + 3 black chips

Total chips=9

n=5

Probability of (Red chips ) can be determined by

=$\frac{6}{9}$

=$\frac{2}{3}$

=0.667

Now we used the binomial theorem

$P(x) = C(n,x)*px*(1-p)(n-x).....Eq(1)\\ putting \ the \ given\ value \ in\ Eq(1)\ we \ get \\p(x=1) = C(5,1) * 0.667^1 * (1-0.667)^4$

This can give 0.0412

Answer 2

Answer:

The probability that one red chip was selected is 0.0053.

Step-by-step explanation:

Let the random variable X be defined as the number of red chips selected.

It is provided that the selections of the n = 5 chips are done with replacement.

This implies that the probability of selecting a red chip remains same for each trial, i.e. p = 6/9 = 2/3.

The color of the chip selected at nth draw is independent of the other selections.

The random variable X thus follows a binomial distribution with parameters n = 5 and p = 2/3.

The probability mass function of X is:

$P(X=x)={5\choose x}\ (\frac{2}{3})^{x}\ (1-\frac{2}{3})^{5-x};\ x=0,1,2...$

Compute the probability that one red chip was selected as follows:

$P(X=1)={5\choose 1}\ (\frac{2}{3})^{1}\ (1-\frac{2}{3})^{5-1}$

                $=5\times\frac{2}{3}\times \frac{1}{625}\\\\=\farc{2}{375}\\\\=0.00533\\\\\approx 0.0053$

Thus, the probability that one red chip was selected is 0.0053.