Probability

Given: P(A) = 0.2, P(B) = 0.55, and P(A ∩ B) = 0.1. We are asked to find P(B ∩ A^c), which is the probability of event B occurring and event A not occurring.

Step 1: Use the formula for P(B):

Step 2: Substitute the given values into the formula:

Step 3: Solve for P(B ∩ A^c):

Final Answer:

Given: An urn contains 25 marbles numbered from 1 to 25. A marble is chosen at random two times with replacement. We are asked to find the probability that both times the marble has the same number.

Step 1: Probability of choosing any particular marble in the first draw:

Step 2: Since the marble is chosen with replacement, the probability of choosing the same marble again in the second draw is also:

Step 3: Total probability of both draws resulting in the same marble:

Final Answer:

Given: A bag contains 5 yellow, 3 green, 2 blue, and 7 white balls. A total of 5 + 3 + 2 + 7 = 17 balls are present.

Step 1: Total ways to choose 4 balls from 17 balls:

Step 2: Ways to choose 4 balls such that none are white (i.e., choosing from 10 non-white balls):

Step 3: Calculate the probability:

Final Answer:

Let the sample space represent all possible outcomes when two dice are rolled. The total number of possible outcomes is . Hence, .

Event A: The sum of the numbers on the two dice is 4. The favorable outcomes are:

• (1, 3)
• (2, 2)
• (3, 1)

So, .

Event B: At least one of the dice shows a 3. The favorable outcomes are:

• (3, 1)
• (3, 2)
• (3, 3)
• (3, 4)
• (3, 5)
• (3, 6)
• (1, 3)
• (2, 3)
• (4, 3)
• (5, 3)
• (6, 3)

So, .

Event A ∩ B: The favorable outcomes where the sum is 4 and at least one die shows a 3 are:

• (1, 3)
• (3, 1)

So, .

Now, we use the conditional probability formula:

Substitute the values:

Answer:

We are given the following information:

• The probability of getting a COVID-19 infection for a vaccinated person is reduced from 0.4 to 0.1.
• 45% of people in the city are vaccinated.
• We need to find the probability that a non-vaccinated person, chosen at random, gets a COVID-19 infection.

Let:

• The probability that a person is vaccinated is 45%, or .
• The probability that a person is non-vaccinated is 55%, or .

We are asked to find the probability that a non-vaccinated person gets COVID-19. Since we are only concerned with non-vaccinated people, the answer is simply the probability of infection for non-vaccinated people:

The probability that a non-vaccinated person chosen at random in the city gets COVID-19 infection is 0.4.

We draw a marble 4 times with replacement from the set {1, 2, 5, 10}. Each draw is independent. Since there are 4 possible outcomes for each of the 4 draws, the total number of possible sequences (outcomes in the sample space) is:
Total Outcomes = 4 × 4 × 4 × 4 = 4⁴ = 256.

Identify Favorable Outcomes (Sum = 18)

We need to find sequences of 4 numbers (n₁, n₂, n₃, n₄) chosen from {1, 2, 5, 10} such that n₁ + n₂ + n₃ + n₄ = 18.

Let's find the combinations of numbers that sum to 18:

• The largest number is 10. If we use 10 twice (10+10=20), the sum is already > 18. So, we can use '10' at most once.
• If we don't use '10', the maximum sum is 5+5+5+5 = 20. Let's try combinations using only {1, 2, 5}:
  • 5, 5, 5, ? -> Sum is 15, need 3 (not in the set).
  • 5, 5, ?, ? -> Sum is 10, need 8 from two numbers in {1, 2, 5}. Possible pairs: (5, 3) - no 3; (2, 6) - no 6; (1, 7) - no 7. Cannot make 8 from {1, 2, 5}.
• So, we must use '10' exactly once. The remaining three numbers must sum to 18 - 10 = 8. These three numbers must be chosen from {1, 2, 5}.
  • Can we use 5? Yes. If one number is 5, the other two must sum to 8 - 5 = 3. From {1, 2, 5}, the only way to get 3 is 1 + 2. So, the three numbers are {5, 2, 1}.
  • Can we avoid using 5? The three numbers must be from {1, 2} and sum to 8. The maximum sum is 2+2+2 = 6. So, we cannot make 8 using only {1, 2}.

Therefore, the only combination of numbers (ignoring order for now) that sums to 18 is {1, 2, 5, 10}.

The favorable outcomes are the sequences formed by arranging the numbers {1, 2, 5, 10}. Since all four numbers are distinct, the number of ways to arrange them in a sequence of 4 draws is the number of permutations of these 4 numbers. Number of Favorable Sequences = 4! = 4 × 3 × 2 × 1 = 24.

The probability is the ratio of the number of favorable sequences to the total number of possible sequences. Probability = (Number of Favorable Sequences) / (Total Number of Outcomes)
Probability =

Simplify the fraction: Divide both numerator and denominator by their greatest common divisor, which is 8.
Probability = = .

The probability that the sum of the numbers equals 18 is 3/32.

The correct option is (D) :

We are given:
- 4 red marbles,
- 3 blue marbles,
- 3 yellow marbles.

Three marbles are drawn simultaneously, and we need to find the probability that the number of yellow marbles drawn will be less than 2. This means we are interested in cases where either 0 or 1 yellow marble is drawn.

Step 1: Total number of ways to draw 3 marbles
The total number of ways to draw 3 marbles from a set of 10 (4 red + 3 blue + 3 yellow) marbles is given by the combination:

Step 2: Number of favorable outcomes
We need to calculate the number of ways to draw 0 or 1 yellow marble.

Case 1: Drawing 0 yellow marbles
If 0 yellow marbles are drawn, we must select all 3 marbles from the red and blue marbles. There are 7 marbles (4 red and 3 blue), and the number of ways to choose 3 marbles from these 7 is:

Case 2: Drawing 1 yellow marble
If 1 yellow marble is drawn, we must select 2 marbles from the 7 red and blue marbles. The number of ways to choose 2 marbles from these 7 is:

Then, there are 3 ways to choose 1 yellow marble from the 3 yellow marbles:

So, the total number of favorable outcomes for this case is:

Step 3: Total favorable outcomes
The total number of favorable outcomes is the sum of the outcomes from Case 1 and Case 2:

Step 4: Calculate the probability
The probability that the number of yellow marbles drawn is less than 2 is the ratio of favorable outcomes to total outcomes:

So, the correct option is (B) :

We are tossing a fair coin twice. Given that the first toss resulted in head, the outcome of the second toss is independent of the first toss. The probability of getting heads on any fair coin toss is . Therefore, the probability that the second toss will also result in heads is simply .

So, the correct option is (D) :

The total number of people at the party is the sum of the number of men and women: The number of women is 33. The probability of choosing a woman is the ratio of the number of women to the total number of people:

The correct option is (A) :

We are given the following probabilities: We need to find , which is the probability of event occurring given that either or occurs. Using the conditional probability formula: Since , we have: Now, we calculate using the formula for the union of two events: Substituting the given values: Thus:

The correct option is (C) :

We are given that three fair dice are rolled simultaneously. Let be the numbers on the top of the dice. We are asked to find the probability that the minimum value among is 6.
Step 1: Understanding the condition For , this means that all three dice must show at least 6, and the smallest value among should be exactly 6. In other words, at least one die must show a 6, and no die should show a value less than 6. Therefore, the only possible outcome for is that one die shows a 6 and the other two dice must show 6 as well.
Step 2: Count the number of favorable outcomes For , all three dice must show 6. There is only one favorable outcome: .
Step 3: Count the total number of outcomes Since each die has 6 faces, the total number of possible outcomes when rolling 3 dice is:
Step 4: Calculate the probability The probability is the ratio of favorable outcomes to the total number of possible outcomes.
Since there is only 1 favorable outcome ( ) out of 216 possible outcomes, the probability is:

The correct option is (A) :

Give: The six dice rolled for many number of times .

Then the outcomes of the event:

for each face the probability is ,

Therefore, the mean value,

For the six-faced dice, the mean value would be:

Hence the mean value of the outcome is 3.5.

Given:

• A biased die where the probability of rolling dots ( ) is proportional to

Step 1: Let the probability of rolling dots be , where is the constant of proportionality.

Step 2: Since probabilities must sum to 1:

Step 3: The probability of rolling a 5 is:

The correct answer is (E) .

Given: Odds in favor of A = 1:1.

Odds in favor of B = 3:2.

Therefore, probability of A P(A) = and probability of B P(B) = .

As per the question we need to find the probability that only one of the two events occurs. This means either event A occurs (but not B), or event B occurs (but not A).

Probability of only A occurring = P(A) × (1 - P(B)) =

Probability of only B occurring = P(B) × (1 - P(A)) =

Now, to find the total probability that only one of the two events occurs, we can write,

Total probability = Probability of only A + Probability of only B

So, the probability that only one of the two events A or B occurs is .

Hence the correct answer is

Given the probability distribution of random variable :

Step 1: Find the value of using the fact that probabilities must sum to 1:

Step 2: Calculate :

Given:

• Sample space
• Events:

We need to find .

Step 1: Find :

Step 2: Find its complement :

Step 3: Find :

Step 4: Compute the union :

Step 5: Calculate the probability:

The correct answer is (D) .

Given:

• A box contains 100 tickets numbered from 00 to 99
• Random variable X = sum of digits on the ticket
• Random variable Y = product of digits on the ticket

Step 1: Find all tickets where (product of digits is 0):

These are tickets where at least one digit is 0: Total of 19 such tickets (00-09 = 10, plus 10-90 in steps of 10 = 9, totaling 19).

Step 2: Find tickets where both and :

These are: But only 02 and 20 satisfy both conditions (11 has product 1 ≠ 0).

Step 3: Calculate the conditional probability:

The correct answer is (D) .

Given:

• Urn contains 8 black marbles and 4 white marbles
• Total marbles = 8 + 4 = 12
• Two marbles are drawn without replacement

Step 1: Calculate the probability of first marble being black:

Step 2: Calculate the probability of second marble being black (after first black is drawn):

Step 3: Multiply the probabilities:

Given:

• Types of bulbs: A, B, C
• Defective rates:
  • 6% of type A are defective
  • 4% of type B are defective
  • 2% of type C are defective
• Quantity in lot:
  • 50 type A bulbs
  • 30 type B bulbs
  • 20 type C bulbs

We need to find the probability that a randomly selected defective bulb is type A.

Step 1: Calculate the number of defective bulbs for each type:

Step 2: Compute total defective bulbs:

Step 3: Apply Bayes' Theorem:

The correct answer is (D) .

Given: Total no of coupons = (labelled as 1,2,.....10)

Three coupons are drawn at random and without replacement.

To find probability of max {X1,X2,X3}

Proceeding according to the question, drawing coupons out of =

Now, let's count the number of favorable outcomes (max{X1, X2, X3} < 7).

For max {X1, X2, X3} to be less than , all three coupons X1, X2, and X3 must have values less than and there are coupons (1 to 6) that fulfills the condition.

So, drawing coupons out of these coupons =

Now , Probability= (⇢ Possible events/Total events) =

Given:

• Random variable follows Binomial distribution

We need to find the values of and .

Step 1: Recall the formulas for Binomial distribution:

Step 2: Substitute into :

Step 3: Find using :

Verification:

The correct values are and .

The correct answer is (D) , .

Let represent heads and represent tails.

The sample space for tossing a coin three times is:

There are 8 equally likely outcomes.

Let be the event that a tail occurs in at least two tosses.

There are 4 outcomes in .

Let be the event that a head occurs on the second toss.

There are 4 outcomes in .

We want to find the probability of getting a head on the second toss given that a tail has occurred in at least two tosses. This is .

Using the conditional probability formula:

is the event that a head occurs on the second toss and a tail occurs in at least two tosses.

There are 2 outcomes in .

Therefore,

The probability of getting a head on the second toss given that a tail has occurred in at least two tosses is .

Given:

• Sample space

We need to find the value of .

Step 1: Express all probabilities in terms of :

Step 2: Apply the total probability rule:

Step 3: Calculate :

The correct answer is (C) .

Given:

• (probability of queue at counter A)
• (probability of queue at counter B)
• (probability of queue at both counters)

We need to find the probability that there is no queue at both counters, which is .

Step 1: Use the probability of union:

Step 2: The probability of no queue at both counters is the complement of the union:

However, let's verify using De Morgan's Law:

The correct answer is (C) 0.25.

Given two independent events E and F with probabilities:

Step 1: Use the formula for probability of union of two independent events:

Step 2: Solve for :

The correct value of is .

Step 1: Law of Total Probability
where:
- -
- , so
Step 2: Solve for
Final Answer:

Step 1: Total Outcomes When rolling two dice, there are: total possible outcomes.
Step 2: Favorable Cases The difference of the numbers is zero when both dice show the same number. The valid cases are: So there are favorable outcomes.
Step 3: Compute Probability
Final Answer: