Ncert Solutions Maths class 11th

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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Ncert Solutions Maths class 11th Permutations and Combinations

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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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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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22. The sample space of the experiment is

S = { (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)}

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

(3, 1), (3, 2), (3, 3), (3, 4

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22. The sample space of the experiment is

S = { (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)}

Now, 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 = { (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)}

C = { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

(i) A? = S – A = Sample space – getting even number on 1st dice = getting odd number on 1st dice = B

(ii) B' = S – B = Sample space – getting odd number on 1st dice = getting even number on 1st dice = A

(iii) A or B = A∪B = (getting even on 1st die)∪ (getting odd o 1st dice) = Sample space = S

(iv) A and B = A∩ B = (getting even on 1st dice)∩ (getting odd on 1st dice) =∅

(v) A but not 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)}

A – C = { (2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(vi) B or C = B∪C = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 1), (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), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

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

(vii) B and C = B∩ C = { (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), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} = { (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2), (4, 1)}

(viii) A∩B ∩C = A ∩A ∩C (as B' = A)

= A ∩C' (A ∩A = A)

So, C' = S – C = { (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), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

C' = { (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (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)}

So, A ∩B ∩ 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, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (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)}

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

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21. Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive.

(iii) Two events, which are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive.

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21. The sample space of the experiment is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

(i) Let A and B be two events such that

A: only head occurs

A = {HHH}

B: only tail occurs

B = {TTT}

And A ∩B = {HHH}∩ {TTT} =∅

(ii) Let A, B and C be two events such that

A: at most one head (i.e., the event in which we get maximum one head; one head or no head at all) occurs

So, A = {HTT, THT, TTH, TTT}

B: exactly two head occurs

So, B = {HHT, HTH, THH}

C: all are heads

So, C = {HHH}

Hence, A ∩ B = {HTT, THT, TTH, TTT} ∩ {HHT, HTH, THH} =∅

A ∩C = {HTT, THT, TTH, TTT} ∩ {HHH} =∅

B∩ C = {HHT, HTH, THH}∩ {HHH} =∅

And A∪ B ∪C = {HTT, T

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20. Three coins are tossed once. Let A denote the event 'three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails

show and D denote the event 'a head shows on the first coin”. Which events are

(i) Mutually exclusive ? (ii) Simple? (iii) Compound?

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20. The sample space of the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Now, A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

(i) A∩ B = {HHH}∩ {HHT, HTH, THH} =∅

A ∩ C = {HHH}∩ {TTT} =∅

A∩ D = {HHH}∩ {HHH, HTH, HTT} = {HHH}

B ∩ C = {HHT, HTH, THH}∩ {TTT} =∅

B∩ D = {HHT, HTH, THH}∩ {HHH, HHT, HTH, HTT} = {HHT, HTH}

C∩ D = {TTT}∩ {HHH, HHT, HTH, HTT} =∅

Hence, (A, B), (A, C), (B, C) and (C, D) are pairs of mutually exclusive events.

(ii) Simple event are those which has only one sample point. So, the simple event are A and C.

(iii) Event having more than one sample point are called com

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19. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

A: the sum is greater than 8, B: 2 occurs on either die

C: the sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?

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