Respuesta :
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:
[tex]P(X=x)={5\choose x}\ (\frac{2}{3})^{x}\ (1-\frac{2}{3})^{5-x};\ x=0,1,2...[/tex]
Compute the probability that one red chip was selected as follows:
[tex]P(X=1)={5\choose 1}\ (\frac{2}{3})^{1}\ (1-\frac{2}{3})^{5-1}[/tex]
[tex]=5\times\frac{2}{3}\times \frac{1}{625}\\\\=\farc{2}{375}\\\\=0.00533\\\\\approx 0.0053[/tex]
Thus, the probability that one red chip was selected is 0.0053.
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
=[tex]\frac{6}{9}[/tex]
=[tex]\frac{2}{3}[/tex]
=0.667
Now we used the binomial theorem
[tex]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[/tex]
This can give 0.0412