Question

A box contains 5 red and 10 white marbles. If 8 marbles are to be chosen at random (without replacement), determine the probability that (a) 4 will be red, (b) all will be white, (c) at least one will be red.

Answer 1

Answer:

a) 0.0023

b) 0.007

c) 0.993

Step-by-step explanation:

The total number of marbles are;

5 + 10 = 15

Probability of red = 5/15

Probability of white = 8/15

Probability of 4 being red means, 4 will be white too;

Thus, the probability without replacement will be;

5/15 * 4/14 * 3/13 * 2/12 * 10/11 * 9/10 * 8/9 * 7/8 =

0.002331002331 which is approximately 0.0023

b. All white

Here, we have all the marbles selected being white

Thus the probability without replacement will be;

10/15 * 9/14 * 8/13 * 7/12 * 6/11 * 5/10 * 4/9 * 3/8

= 0.006993006993 which is approximately 0.007

c. At least one will be red

Mathematically, that will be;

1 - p(no red)

p(no red) = p(all white)

P(all white) = 0.007

hence P(no red) = 0.007

P(at least one red) = 1 - 0.007 = 0.993

Answer 2

Using the hypergeometric distribution, it is found that there is a:

a) 0.1632 = 16.32% probability that 4 will be red.

b) 0.0070 = 0.7% probability that all will be white.

c) 0.993 = 99.3% probability that at least one will be red.

The marbles are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• N is the size of the population.
• n is the size of the sample.
• k is the total number of desired outcomes.

In this problem:

• There are 15 marbles, hence $N = 15$
• There are 5 red marbles, hence $k = 5$.
• There are 8 marbles chosen, hence $n = 8$

Item a:

This is P(X = 4), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,15,8,5) = \frac{C_{5,4}C_{10,4}}{C_{15,8}} = 0.1632$

0.1632 = 16.32% probability that 4 will be red.

Item b:

0 red, hence P(X = 0).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,15,8,5) = \frac{C_{5,0}C_{10,8}}{C_{15,8}} = 0.0070$

0.0070 = 0.7% probability that all will be white.

Item c:

This is:

$P(X \geq 1) = 1 - P(X = 0)$

From item b:

$P(X \geq 1) = 1 - 0.007 = 0.993$

0.993 = 99.3% probability that at least one will be red.

A similar problem is given at https://brainly.com/question/24320200