حل اسئلة احصاء واحتمالية قسم الحاسوب الجامعة المستنصرية نموذج رقم 1

Q1: A: How many "numbers" of three digit and more than "400" can be formed from the digits

{1,2,3,4,5,6}.

Sol//

to find the number of three-digit numbers that can be formed using the digits {1, 2, 3, 4, 5, 6} and greater than 400, we need to consider the possible combinations based on the conditions given.

Since the number needs to be greater than 400, the first digit must be 4 or greater. For the second and third digits, we have all six digits to choose from.

Let's break it down:

First digit: It can be 4, 5, or 6 (3 possibilities).
Second digit: Any of the six digits {1, 2, 3, 4, 5, 6} (6 possibilities).
Third digit: Any of the six digits {1, 2, 3, 4, 5, 6} (6 possibilities).

Using the multiplication principle, the total number of three-digit numbers greater than 400 that can be formed is:

$\text{Total}=\text{Number of possibilities for the first digit}\times \text{Number of possibilities for the second digit}\times \text{Number of possibilities for the third digit}$

$\text{Total}=3\times 6\times 6$

$\text{Total}=108$

So, there are 108 three-digit numbers that can be formed from the digits {1, 2, 3, 4, 5, 6} and are greater than 400.

B: For the experiment of rolling 2 dices, what the probability of at least one of the two dices is

3?

Sol//

To find the probability of at least one of the two dice showing a 3 when rolled, we can use the principle of complementary probability.

First, let's find the probability of the complement event, which is the probability of neither die showing a 3.

Each die has 6 possible outcomes, so there are $6\times 6=36$ total outcomes when rolling two dice.

The number of outcomes where neither die shows a 3 is the number of outcomes when both dice show a number other than 3. Since each die has 5 outcomes other than 3, the number of outcomes where neither die shows a 3 is $5\times 5=25$.

Now, the probability of neither die showing a 3 is:

$𝑃(\text{neither die shows 3})=\frac{\text{number of outcomes where neither die shows 3}}{\text{total number of outcomes}}=\frac{25}{36}$

Since this is the complement event, the probability of at least one die showing a 3 is:

$𝑃(\text{at least one die shows 3})=1-𝑃(\text{neither die shows 3})=1-\frac{25}{36}=\frac{11}{36}$

So, the probability of at least one of the two dice showing a 3 is $\frac{11}{36}$.

O2:A: Suppose that S={a1,a2,a3,a4}, then find P(a1) if you know P(a2)= 1/3 , P(a3)= 1/6,

P(a4)= 1/9.

Sol//

Given:

• $𝑃({𝑎}_2)=\frac{1}{3}$
• $𝑃({𝑎}_3)=\frac{1}{6}$
• $𝑃({𝑎}_4)=\frac{1}{9}$

We know that the sum of probabilities of all outcomes in a sample space is 1. Therefore,

$𝑃({𝑎}_1)+𝑃({𝑎}_2)+𝑃({𝑎}_3)+𝑃({𝑎}_4)=1$

We substitute the known probabilities:

$𝑃({𝑎}_1)+\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=1$

$𝑃({𝑎}_1)+\frac{6}{18}+\frac{3}{18}+\frac{2}{18}=1$

$𝑃({𝑎}_1)+\frac{11}{18}=1$

$𝑃({𝑎}_1)=1-\frac{11}{18}=\frac{18}{18}-\frac{11}{18}=\frac{7}{18}$

So, $𝑃({𝑎}_1)=\frac{7}{18}$.

B: The production of 3 machines A;B and C are 0:50; 0:30 and 0:20, respectively. The rate of

bad production are 0:03; 0:04 and 0:05, respectively. If we choose one material of the

production randomly, what is the probability of this material to be bad?

Sol//

Given:

• $𝑃(𝐴)=0.50$
• $𝑃(𝐵)=0.30$
• $𝑃(𝐶)=0.20$
• $𝑃(𝐷\mathrm{\mid }𝐴)=0.03$
• $𝑃(𝐷\mathrm{\mid }𝐵)=0.04$
• $𝑃(𝐷\mathrm{\mid }𝐶)=0.05$

Substituting the values:

$𝑃(𝐷)=(0.03\times 0.50)+(0.04\times 0.30)+(0.05\times 0.20)$ $𝑃(𝐷)=0.015+0.012+0.010$ $𝑃(𝐷)=0.037$

Therefore, the probability of selecting a defective product randomly is $0.037$ or $3.7\mathrm{\%}$.

Q3: A: Prove or disprove the following statement: If A, B and C are pairwise independence events, then  P(A∩B∩C)= P(A)*P(B)*P(C).

Sol//

The statement "If $𝐴$, $𝐵$, and $𝐶$ are pairwise independent events, then $𝑃(𝐴\cap 𝐵\cap 𝐶)=𝑃(𝐴)\cdot 𝑃(𝐵)\cdot 𝑃(𝐶)$" is not necessarily true.

Pairwise independence means that any pair of events is independent, but it does not imply independence among all three events simultaneously.

To disprove the statement, we can provide a counterexample:

Let's consider three events $𝐴$, $𝐵$, and $𝐶$ such that:

• $𝑃(𝐴)=𝑃(𝐵)=𝑃(𝐶)=\frac{1}{2}$ (each event has probability $\frac{1}{2}$