Question

A person has agreed to participate in an extrasensory perception​ (ESP) experiment. He is asked to randomly pick two numbers between 0 and​ 9, inclusive. The second number must be different from the first. Let H equals event the first number is 4​, and K equals event the second number picked exceeds 8 . Complete parts​ (a) through​ (e) below.a. Determine​ P(H).b. Determine​ P(K |​ H).c. Determine​ P(H &​ K).d. Find the probability that both numbers picked are less than 6e. Find the probability that both numbers picked are greater than 6

Answer 1

Answer:

(a) The probability of H is 0.10.

(b) The value of P (K|H) is 0.10.

(c) The value of P (H and K) is 0.01.

(d) The probability that both the numbers are less than 6 is 0.333.

(e) The probability that both the numbers are more than 6 is 0.067.

Step-by-step explanation:

The sample space of the experiment is:

S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Total number of samples, N = 10.

The events are defined as:

H = the first number is 4.

K = the second number picked exceeds 8.

It is said that both the numbers are different from each other.

This implies that the numbers are selected without replacement.

(a)

The event H is defined the first number is 4.

Compute the probability of H as follows:

$P(H)=\frac{Favorable\ outcomes}{N} =\frac{1}{10} =0.10$

Thus, the probability of H is 0.10.

(b)

The conditional probability of event B given that event A has already happened, provided that A and B are independent is:

$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(A)\times P(B)}{P(A)} =P(B)$

Compute the value of P (K|H) when K and H are independent is:

$P(K|H)=P(K)=\frac{Favorable\ outcomes}{N}= \frac{1}{10}=0.10$

Thus, the value of P (K|H) is 0.10.

(c)

The probability of A and B when events A and B are independent is:

$P(A\cap B)=P(A)\times P(B)$

Compute the value of P (H and K) as follows:

$P(H\cap K) = P(H)\times P(K)=0.10\times0.10=0.01$

Thus, the value of P (H and K) is 0.01.

(d)

There are 10 values in the sample space.

Let X = value is less than 6.

The sample space of X can be defined as:

(X) = {0, 1, 2, 3, 4, 5}

n (X) = 6.

Compute the probability that both the numbers are less than 6 as follows:

P (X) = P (1st number less than 6) × P (2nd number less than 6)

        $=\frac{{6\choose 1}}{{10\choose 1}} \times\frac{{5\choose 1}}{{9\choose 1} }\\=\frac{6}{10}\times\frac{5}{9} \\ =0.333$

Thus, the probability that both the numbers are less than 6 is 0.333.

(e)

Let Y = value is more than 6.

The sample space of X can be defined as:

(Y) = {7, 8, 9}

n (Y) = 3.

Compute the probability that both the numbers are more than 6 as follows:

P (Y) = P (1st number more than 6) × P (2nd number more than 6)

        $=\frac{{3\choose 1}}{{10\choose 1}} \times\frac{{2\choose 1}}{{9\choose 1} }\\=\frac{3}{10}\times\frac{2}{9} \\ =0.067$

Thus, the probability that both the numbers are more than 6 is 0.067.