Maths

6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

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6. To generate a signal which requires 2 flags one below another we can have the following combination of any of the 5 flag at top and the one of the remaining 4 flag at the bottom.

Hence, total no. of possible combination = 5 * 4 = 20

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25. A coin is tossed twice, what is the probability that atleast one tail occurs?

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25. When a coin is tossed twice we have the sample space

S = {TT, TH, HT, HH}

So, n (S) = 4

Let A be the event of getting at least one tail.

Then, A = {TH, HT, TT}

So, n (A) = 3

Therefore, P (A)= Numver of outcome favorable to A/Total possible outcomes = $\frac{}{} = \frac{n (A)}{n (S)}$ .

$= \frac{3}{4}$

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5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

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5. When a coin is tossed one time a head or a tail is the possible outcome.

So, when a coin is tossed 3 times the total number of possible outcomes = 2 * 2 * 2 = 8

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24. Which of the following can not be valid assignment of probabilities for outcomes

of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇,}

Assignment ω ₁ ω ₂ ω ₃ ω ₄ ω ₅ ω ₆ ω ₇

(a) 0.1 &nbs

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24. Which of the following can not be valid assignment of probabilities for outcomes

of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇,}

Assignment ω ₁ ω ₂ ω ₃ ω ₄ ω ₅ ω ₆ ω ₇

(a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6

(b)

(c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7

(d) – 0.1 0.2 0.3 0.4 – 0.2 0.1 0.3

(e) $\frac{\bar{1}}{\bar{14}}$ $\frac{\bar{2}}{\bar{14}}$ $\frac{\bar{3}}{\bar{14}}$ $\frac{\bar{4}}{\bar{14}}$ $\frac{\bar{5}}{\bar{14}}$ $\frac{\bar{6}}{\bar{14}}$ $\frac{\bar{15}}{\bar{14}}$

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24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is 'one' the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

$= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = \frac{7}{7} = 1$ .

P (S) = 1

Hence, the given assignment of probability is valid.

(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7

= 2.8

i.e., P (S) > 1

As probability of the sample space S should always be '1'. The given assignment is invalid.

(d) Here P (W1) = –0.1 is negative.

As p

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24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is 'one' the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

$= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = \frac{7}{7} = 1$ .

P (S) = 1

Hence, the given assignment of probability is valid.

(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7

= 2.8

i.e., P (S) > 1

As probability of the sample space S should always be '1'. The given assignment is invalid.

(d) Here P (W1) = –0.1 is negative.

As probability of happening an event should always be positive and less than or equal to 1.

The given assignment is invalid.

(e) As P (W7) = $\frac{15}{14} = 1.07$

i.e. probability of happening an event is greater than one. The given assignment is invalid.

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4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

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4. For the 5-digit telephone number that can be constructed using 0 to 9 if each number starts with 67 and no digit appears more than once we can have

6 numbers

7 numbers

8 numbers

So total number of possible combination = 1 * 1 * 6 * 7 * 8 = 336

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3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

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3. To form a 4-letter code using the first 10 letters of the English alphabet without repeating we can have 10, 9, 8 and 7 numbers of letters to be filled at ones, tens, hundreds and thousands place simultaneously.

Hence, total no. of 4-letter code that can be made using the first 10-letter of English alphabet = 7 * 8 * 9 * 10 = 5040

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2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

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Maths Permutation and Combination

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2. Since we are given with six numbers (1, 2, 3, 4, 5, 6) to form 3-digit even number and also repetition is allowed.

We can have only the number 2, 4, 6 in the ones place and all the six numbers can fill the tens and hundreds place.

So, total number of 3-digit even number that can be formed by 1, 2, 3, 4, 5, 6

= 6 * 6 * 3

= 108

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Permutation and Combination Exercise 6.1 Solutions

1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed?

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1. i. Since repetition of number is allowed and there are five numbers which can be used to form the necessary 3-digit numbers we can have five numbers that can fill the ones, tens and hundreds place.

So, total number of possible 3-digit number = 5 * 5 * 5 = 125

ii. Since repetition of number is not allowed. There are total 5 numbers which can fill the ones places then 4 and 3 numbers which can fill the tens and hundreds place simultaneously.

So, total number of possible 3-digit number = 3 * 4 * 5 = 60

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23. Refer to question 6 above, state true or false: (give reason for your answer)

(i) A and B are mutually exclusive

(ii) A and B are mutually exclusive and exhaustive

(iii) A = B′

(iv) A and C are mutually exclusive

(v) A and B′ are mutually exclusive.

(vi) A′, B′, C are mutually exclusive and exhaustive.

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23. (i) A ∩B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

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

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

A ∩B =∅

Hence, A and B are mutually exclusive.

The given statement is true.

(ii) A ∪B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1,

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23. (i) A ∩B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

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

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

A ∩B =∅

Hence, A and B are mutually exclusive.

The given statement is true.

(ii) A ∪B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

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

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

= { (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)}

= S

Hence, A B = S and A∩ B =∅

A and B are mutually exclusive and mutually exhaustive.

So, the give statement is true.

(iii) A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

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

B = S – B = { (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)} – { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, (4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

= { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

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

B' = A

Hence, the given relation is true.

(iv) A ∩C = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

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

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, (1)}

= { (2, 1), (2, 2), (2, 3), (4, 1)}≠∅

So, A∩ C≠∅

i.e., A and C are not mutually exclusive.

Hence, the given statement is false.

(v) A∩ B' = A∩ A = A≠∅ { B' = A}

Hence A∩B'≠∅ i.e., A and B are not mutually exclusive.

So, the given statement is false.

(vi) As A' = B

B = A

So, A'∩B = B∩A =∅

A'∩ C = B∩C = { (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}=∅

B'∩C = A∩C = { (2, 1), (2, 2), (2, 3), (4, 1)}=∅

Hence, A', B' and C are not mutually exclusive.

The given statement is false.

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22. Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5.

Describe the events

(i) A′ (ii) Not B (iii) A or B

(iv) A and B (v) A but not C

(vi) B or C (vii) B and C (viii) A ∩ B′ ∩ C′

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