حل اسئلة احصاء واحتمالية قسم الحاسوب الجامعة المستنصرية نموذج رقم 1

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 حل اسئلة احصاء واحتمالية قسم الحاسوب الجامعة المستنصرية نموذج رقم 1

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:

Total=Number of possibilities for the first digit×Number of possibilities for the second digit×Number of possibilities for the third digit

Total=3×6×6

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 6×6=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 5×5=25.

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

𝑃(neither die shows 3)=number of outcomes where neither die shows 3total number of outcomes=2536

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

𝑃(at least one die shows 3)=1𝑃(neither die shows 3)=12536=1136

So, the probability of at least one of the two dice showing a 3 is 1136.


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(𝑎2)=13
  • 𝑃(𝑎3)=16
  • 𝑃(𝑎4)=19

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

𝑃(𝑎1)+𝑃(𝑎2)+𝑃(𝑎3)+𝑃(𝑎4)=1

We substitute the known probabilities:

𝑃(𝑎1)+13+16+19=1

𝑃(𝑎1)+618+318+218=1

𝑃(𝑎1)+1118=1

𝑃(𝑎1)=11118=18181118=718

So, 𝑃(𝑎1)=718.


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(𝐷𝐴)=0.03
  • 𝑃(𝐷𝐵)=0.04
  • 𝑃(𝐷𝐶)=0.05

Substituting the values:

𝑃(𝐷)=(0.03×0.50)+(0.04×0.30)+(0.05×0.20) 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%.

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

Sol//

The statement "If 𝐴, 𝐵, and 𝐶 are pairwise independent events, then 𝑃(𝐴𝐵𝐶)=𝑃(𝐴)𝑃(𝐵)𝑃(𝐶)