Probability Exercise 04
Question:
Two numbers are selected at random (without replacement) the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Answer:

The two numbers can be selected from the set {1, 2, 3, 4, 5, 6}.

Since the numbers are selected at random and without replacement, the probability of selecting any two numbers is equal.

Let X denote the larger of the two numbers obtained. The possible values of X are 2, 3, 4, 5, and 6.

The expected value of X is given by the formula E(X) = ∑xP(x), where x is the possible value of X and P(x) is the probability of obtaining that value.

To calculate the probability of obtaining a particular value, we need to consider all the possible combinations of two numbers from the set {1, 2, 3, 4, 5, 6}.

There are 15 possible combinations of two numbers from the set {1, 2, 3, 4, 5, 6}.

The probability of obtaining a particular value of X is given by the number of combinations in which that value of X is obtained divided by the total number of combinations.

For example, the probability of obtaining X = 2 is 3/15, since there are three combinations in which X = 2 (i.e., (1,2), (2,1), (2,2)).

Similarly, the probability of obtaining X = 3 is 4/15, X = 4 is 3/15, X = 5 is 2/15, and X = 6 is 1/15.

Thus, the expected value of X is given by E(X) = ∑xP(x) = (23/15) + (34/15) + (43/15) + (52/15) + (6*1/15) = 3.4
Question:
Find the probability distribution of (i) number of heads in two tosses of a coin.
Answer:
(i) The probability distribution of the number of heads in two tosses of a coin is given by the following table:
Number of Heads Probability 0 1/4 1 2/4 2 1/4
Question:
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Answer:
Step 1: Calculate the total number of outcomes when two dice are thrown simultaneously.
There are 36 possible outcomes when two dice are thrown simultaneously.
Step 2: Calculate the number of outcomes when two dice show a six.
There are 4 possible outcomes when two dice show a six (66, 61, 16, 66).
Step 3: Calculate the probability of getting two sixes when two dice are thrown simultaneously.
The probability of getting two sixes when two dice are thrown simultaneously is 1/36.
Step 4: Calculate the expectation of X.
The expectation of X = (4*1/36) = 1/9
Question:
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X ? Find mean, variance and standard deviation of X.
Answer:
Probability Distribution of X:
X P(X) 14
Question:
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer:

Let X be the random variable representing the number of tails.

X can take values 0, 1, or 2.

The probability of getting 0 tails is (3/4)^{2} = 9/16.

The probability of getting 1 tail is 2(3/4)(1/4) = 3/8.

The probability of getting 2 tails is (1/4)^{2} = 1/16.

Therefore, the probability distribution of X is:
X=0: 9/16
X=1: 3/8
X=2: 1/16
Question:
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ? Is X a random variable ? A 0,1,2 B 3,5,7 C 7,7,8 D 1,5,7
Answer:
A 0,1,2; Yes, X is a random variable.
Question:
The random variable X has a probability distribution P(X) of the following form, where k is some number : P(X)={k,ifx=02k,ifx=13k,ifx=20,otherwise} (a) Determine the value of k (b) Find P(X<2),P(X≤2),P(X≥2).
Answer:
(a) To determine the value of k, we need to add up the probabilities for each of the values of X, which is 0, 1, and 3. This gives us k + 2k + 3k = 1, so k = 1/6.
(b) P(X<2) = P(X=0) = k = 1/6. P(X≤2) = P(X=0) + P(X=1) = k + 2k = 1/2. P(X≥2) = P(X=3) = 3k = 1/2.
Question:
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is: A 1 B 2 C 5 D 38
Answer:
Answer: B 2
Question:
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer:

Let X be the random variable representing the number of defective bulbs.

The possible values of X are 0, 1, 2, 3, and 4.

The probability of drawing 0 defective bulbs is (24/30)^{4} = 0.25.

The probability of drawing 1 defective bulb is (6/30) * (24/30)^{3} = 0.3.

The probability of drawing 2 defective bulbs is (6/30)^{2} * (24/30)^{2} = 0.2.

The probability of drawing 3 defective bulbs is (6/30)^{3} * (24/30) = 0.15.

The probability of drawing 4 defective bulbs is (6/30)^{4} = 0.03.

The probability distribution of the number of defective bulbs is given by:
X = 0, P(X = 0) = 0.25
X = 1, P(X = 1) = 0.3
X = 2, P(X = 2) = 0.2
X = 3, P(X = 3) = 0.15
X = 4, P(X = 4) = 0.03
Question:
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X=0 if he opposed, and X=1 if he is in favour. Find E(X) and Var(X).
Answer:
E(X) = 00.3 + 10.7 = 0.7
Var(X) = (00.7)^{2}*0.3 + (10.7)^{2}*0.7 = 0.21
Question:
Find the mean number of heads in three tosses of a fair coin.
Answer:
Step 1: Determine the possible outcomes of tossing a fair coin three times.
Answer: The possible outcomes of tossing a fair coin three times are: headsheadsheads, headsheadstails, headstailstails, tailstailstails, tailstailsheads, tailsheadsheads.
Step 2: Count the total number of possible outcomes.
Answer: There are six possible outcomes of tossing a fair coin three times.
Step 3: Count the number of outcomes that result in heads.
Answer: There are four outcomes that result in heads: headsheadsheads, headsheadstails, headstailstails, tailsheadsheads.
Step 4: Calculate the mean number of heads.
Answer: The mean number of heads is 4/6, or 2/3.
Question:
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is A 37/221 B 5/13 C 1/13 D 2/13
Answer:
Answer: D 2/13
Question:
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 (ii) six appears on at least one die
Answer:
Answer:
Step 1: List out all the possible outcomes of two tosses of a die.
Outcomes = (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Step 2: Count the number of successes. A success is defined as number greater than 4 or six appearing on at least one die.
Number of successes = 11 (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (4,6), (5,5), (5,6), (6,6)
Step 3: Calculate the probability of each outcome.
Total number of outcomes = 36
Probability of success = 11/36
Step 4: Construct the probability distribution table.
Outcome
Question:
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X ? 9,7,4,0 0,2,4,6 6,7,7,2 6,4,2,0
Answer:
Answer: 0,2,4,6
Question:
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Answer:
Step 1: Calculate the expected value of X.
Expected value of X = (1 + 2 + 3 + 4 + 5 + 6) + (1 + 2 + 3 + 4 + 5 + 6) = 7 + 7 = 14
Step 2: Calculate the variance of X.
Variance of X = [ (1  14)2 + (2  14)2 + (3  14)2 + (4  14)2 + (5  14)2 + (6  14)2 + (7  14)2 + (8  14)2 + (9  14)2 + (10  14)2 + (11  14)2 + (12  14)2 ] / 36 = [49 + 36 + 25 + 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16] / 36 = 205 / 36 = 5.69444
Step 3: Calculate the standard deviation of X.
Standard deviation of X = √5.69444 = 2.3818