Solution

For n = 2: E: The only possibility is having one boy and one girl, so P(E) = 1. F: The possibilities are having two girls or one boy and one girl, so P(F) = 1/2. Since P(E) * P(F) = 1 * 1/2 = 1/2 ≠ P(E ∩ F) = 0, E and F are not independent for n = 2.

For n = 4: E: The possibilities are having two boys and two girls, three boys and one girl, or one boy and three girls, so P(E) = 3/8. F: The possibilities are having four girls, three girls and one boy, two girls and two boys, or one girl and three boys, so P(F) = 4/16 = 1/4. Since P(E) * P(F) = 3/8 * 1/4 = 3/32 ≠ P(E ∩ F) = 0, E and F are not independent for n = 4.

For n = 5: E: The possibilities are having two boys and three girls, three boys and two girls, four boys and one girl, or five boys, so P(E) = 4/16 = 1/4. F: The possibilities are having five girls, four girls and one boy, three girls and two boys, two girls and three boys, or one girl and four boys, so P(F) = 5/32. Since P(E) * P(F) = 1/4 * 5/32 = 5/128 ≠ P(E ∩ F) = 0, E and F are not independent for n = 5.

Solution

Solution

Solution

1. Start with the universe: 26!
2. Subtract the permutations that contain:

• fish.

To count the number of permutations with a fixed substring:

Treat the fixed substring as 1 letter.

There are (26 – 4 + 1)! such permutations.

• rat: (26 – 3 + 1)!
• bird: : (26 – 4 + 1)!

1. Add the permutations that contain:

• fish & rat: (26 – 4 – 3 + 2)!
• fish & bird: 0
• rat & bird: 0

1. Subtract the permutations that contain all 3 strings

There are 0 such permutations.

Answer: 26! – 23! – 24! – 23! + 21!

Solution

Solution

Solution

Using combinations with repetition, the formula to calculate the number of ways to choose r items from n types of items with unlimited supply is given by the formula:

C(n+r-1, r)

In this case, we have 3 types of fruits (apples, pears, apricots) and we want to choose 5 fruits in total.

So, the number of ways to choose 5 fruits from the market stall using combinations with repetition is:

C(3+5-1, 5) = C(7, 5) = 21 ways

Therefore, the correct answer using combinations with repetition is 21 ways.

Solution

This is calculated by considering all possible combinations of signals A and B that add up to 10 seconds. Signal A lasts for 1 second and signal B lasts for 2 seconds. So, we can have combinations like AAAABBBBB, AABAAABBB, BBAAAAAAB, and so on. The total number of such combinations is 89.

Solution

To find the conditional probability P(G|F), we use the formula:

P(G|F) = P(G ∩ F) / P(F)

Since G and F are independent events (disjoint events are a special case of independent events), P(G ∩ F) = P(G) * P(F) = 0.25 * 0.25 = 0.0625.

Therefore, P(G|F) = 0.0625 / 0.25 = 0.25

Solution

Solution

Solution

Case 1: No bit in error:

The probability of no bit being in error in a block of n bits is (1-p) for each bit. Since the bits are independent, the probability of all n bits being error-free is (1-p)^n.

Case 2: Exactly one bit in error:

The probability of exactly one bit being in error can occur in any one of the n positions in the block. So, we need to multiply the probability of a bit being in error (p) with the probability of all other bits being error-free (1-p)^(n-1). Therefore, the probability of exactly one bit being in error is n * p * (1-p)^(n-1).

Total probability:

To find the probability of at most one bit in error, we need to add the probabilities of Case 1 and Case 2: P(at most one bit in error) = (1-p)^n + n * p * (1-p)^(n-1)

Simplifying this expression, we get:

P(at most one bit in error) = (1-p)^(n-1) * [(1-p) + n * p]

Comparing the simplified expression with the given options, we can see that option C matches the form of the expression: np(1-p)^(n-1

Solution

Solution

Solution

Let us choose first the value for the three cards, which gives 13 choices, and there are

4C3 configurations of suits. For the pair, there are 12 possible values and 4C2 suits.

So, number of full house possible

Solution

Exactly 4 in correct place = 6C4 * D2 = 15

Exactly 5 in correct place = 6C5 * D1 = 6

Exactly 6 in correct place = 1

Total no of ways = 22

Solution

5 letters be put into 3 letter boxes A,B,C…

means 5 different letters in 3 different boxes..

condition : letter box A contains at least 2 letters…

required ways where letter box A has at least 2 letters= box A has 2 letters + box A has #letters + box A has 4 letters + box A has 5 letters

=5C2*2^3 + 5C3*2^2+5C4*2^1 +5C5

=80+40+10+1 =131 answer

Solution

Solution

First find E(X):

Now find E(X^2 ):

Now find V(X):

Solution

Solution

Solution

(b)

(c)

(d)

Solution

= 0.333

Solution

Option A:

Option B

Option C

Option D

Solution

Solution

Solution

Solution

Solution

We can distribute the books in 5! ways the first week. Numbering both the books and the reviewers (for the first week) as 1, 2,..., 5, for the second distribution we must arrange these numbers so that none of them is in its natural position. This we can do in D5 ways. By the rule of product, she can make the two distributions in (5!) D5 = 120 * 44 = 5280

Solution

Solution

The possible values of X are 1, 2, 5 and 10. We have to find the probabilities of these outcomes, that is, the probability mass function of X. It helps if we introduce another random variable: let R € [0, 9] denote the distance of the dart to the center.

Then for any a € [0,9] we have

Solution

Solution