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 $\frac{1}{2}$)
• $𝑃(𝐴\cap 𝐵)=𝑃(𝐵\cap 𝐶)=𝑃(𝐴\cap 𝐶)=0$ (each pair of events is mutually exclusive)

Now, let's calculate $𝑃(𝐴\cap 𝐵\cap 𝐶)$:

$𝑃(𝐴\cap 𝐵\cap 𝐶)=𝑃(\mathrm{\varnothing })=0$

However,

$𝑃(𝐴)\cdot 𝑃(𝐵)\cdot 𝑃(𝐶)=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}\mathrm{\ne }0$

Therefore, the statement is disproved. Pairwise independence does not guarantee independence among all three events simultaneously.

B: For the experiment of tossing a coin 3 times, let A is the event when the first tossing is H, B

is the event when the second tossing is H and C is the event when we get exactly two

consecutive H. Show that A, B and C are pairwise independent or not.

Sol//

To determine if events $𝐴$, $𝐵$, and $𝐶$ are pairwise independent, we need to check if any two of these events are independent of each other.

Two events $𝑋$ and $𝑌$ are independent if and only if:

$𝑃(𝑋\cap 𝑌)=𝑃(𝑋)\times 𝑃(𝑌)$

Let's evaluate the pairwise independence of events $𝐴$, $𝐵$, and $𝐶$:

1. Pairwise independence of $𝐴$ and $𝐵$: $𝑃(𝐴\cap 𝐵)=𝑃(\text{first toss is H and second toss is H})$ $𝑃(𝐴\cap 𝐵)=𝑃(\text{HH})=𝑃(\text{H})\times 𝑃(\text{H})=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

$𝑃(𝐴)=𝑃(\text{first toss is H})=𝑃(\text{H})=\frac{1}{2}$ $𝑃(𝐵)=𝑃(\text{second toss is H})=𝑃(\text{H})=\frac{1}{2}$

Since $𝑃(𝐴\cap 𝐵)=𝑃(𝐴)\times 𝑃(𝐵)$, events $𝐴$ and $𝐵$ are independent.

2. Pairwise independence of $𝐴$ and $𝐶$: $𝑃(𝐴\cap 𝐶)=𝑃(\text{first toss is H and exactly two consecutive H})$ $𝑃(𝐴\cap 𝐶)=0$

$𝑃(𝐴)=𝑃(\text{first toss is H})=𝑃(\text{H})=\frac{1}{2}$ $𝑃(𝐶)=𝑃(\text{exactly two consecutive H})$

Since $𝑃(𝐴\cap 𝐶)\mathrm{\ne }𝑃(𝐴)\times 𝑃(𝐶)$, events $𝐴$ and $𝐶$ are not independent.

3. Pairwise independence of $𝐵$ and $𝐶$: $𝑃(𝐵\cap 𝐶)=𝑃(\text{second toss is H and exactly two consecutive H})$ $𝑃(𝐵\cap 𝐶)=0$

$𝑃(𝐵)=𝑃(\text{second toss is H})=𝑃(\text{H})=\frac{1}{2}$ $𝑃(𝐶)=𝑃(\text{exactly two consecutive H})$

Since $𝑃(𝐵\cap 𝐶)\mathrm{\ne }𝑃(𝐵)\times 𝑃(𝐶)$, events $𝐵$ and $𝐶$ are not independent.

Q4: A: The sample space for the experiment of tossing a dice once is S={H,T} , find the sigma Field.

Sol//

Therefore, the sigma field $\mathcal{𝐹}$ for the sample space $𝑆=\{𝐻,𝑇\}$ is:

$\mathcal{𝐹}=\{\mathrm{\varnothing },\{𝐻\},\{𝑇\},𝑆\}$

B: How many words of 3 distinct letters can be formed from the letters of the word

“MAST”?

Sol//

In the word "MAST", there are 4 distinct letters, so $𝑛=4$. We want to form words of 3 distinct letters, so $𝑟=3$.

Substituting these values into the permutation formula:

$𝑃(4,3)=\frac{4!}{(4-3)!}$ $=\frac{4!}{1!}$ $=\frac{4\times 3\times 2\times 1}{1}$ $=24$

Therefore, there are $24$ words of 3 distinct letters that can be formed from the letters of the word "MAST".

Q5: Answer only (two) of the following:

A: Prove that P(A∪B)≤P(A)⋅P(B) .

sol//

The statement $𝑃(𝐴\cup 𝐵)\le 𝑃(𝐴)\cdot 𝑃(𝐵)$ is generally false. In fact, $𝑃(𝐴\cup 𝐵)$ is usually greater than $𝑃(𝐴)\cdot 𝑃(𝐵)$ due to the presence of common transitions between events A and B.

Let's provide a simple example:

• Suppose we have a sample set S containing 100 elements.
• Assume we have events A and B, where:
  • $𝑃(𝐴)=0.6$, meaning 60% of the elements in S belong to A.
  • $𝑃(𝐵)=0.7$, meaning 70% of the elements in S belong to B.
  • Let's assume we have common transitions between A and B of 0.3 (meaning 30%).
• In this case:
  • $𝑃(𝐴\cup 𝐵)=𝑃(𝐴)+𝑃(𝐵)-𝑃(𝐴\cap 𝐵)=0.6+0.7-0.3=1.0$ (since the result cannot exceed 1).
  • $𝑃(𝐴)\cdot 𝑃(𝐵)=0.6\times 0.7=0.42$ (as common transitions here are considered an increase in transitions, not a deduction).

Thus, in this example, we find that $𝑃(𝐴\cup 𝐵)$ is greater than $𝑃(𝐴)\cdot 𝑃(𝐵)$.

Therefore, the statement $𝑃(𝐴\cup 𝐵)\le 𝑃(𝐴)\cdot 𝑃(𝐵)$ is generally false.

B: What is the probability of getting the number 5 for the experiment of rolling a dice once?

Sol//

In a standard six-sided die, each face has an equal probability of occurring. Since there is only one face with the number 5, the probability of rolling a 5 on a single roll of a six-sided die is:

$𝑃(5)=\frac{1}{6}$

So, the probability of getting the number 5 for the experiment of rolling a dice once is $\frac{1}{6}$.

C: In how many ways can the letters of English alphabet be arrange so that there are exactly

10 letters between A and Z.

Sol//

Since there are 26 letters in the English alphabet, and we are fixing 12 positions (1 for A, 10 in between, and 1 for Z), we are left with $26-12=14$ remaining letters to arrange in the remaining positions.

The number of ways to arrange 14 distinct objects in the remaining positions is given by the factorial of 14, denoted as $14!$.

تعديل المقال