Maths

28. A fair coin with 1 marked on one face and 6 on the other and a fair die are bothtossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

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28. The sample space of the experiment is

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

So, n (S) = 12.

(i) Let E be event such that sum of numbers that turn up is 3. Then,

E = { (1, 2)}

So, n (E) = 1

P (E) = $\frac{n (E)}{n (S)} = \frac{1}{12}$ .

(ii) Let F be event such that sum of number than turn up is 12. Then,

F = { (6, 6)}

So, n (F) = 1

P (F) = $\frac{n (F)}{n (S)} = \frac{1}{12}$ .

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9. Compute $\frac{8!}{6! * 2!}$

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9. $\frac{8!}{6! * 2!}$ = $\frac{8 * 7 * (6!)}{(6!) * 1 * 2}$ = 4 * 7 = 28

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27. A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(c) Calculate the probability that the card is (i) an ace (ii) black card.

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27. (a) Since there are 52 cards in the sample space,

n (S) = 52.

So, there are 52 sample points.

(b) In a deck of 52 cards there are 4 ace cards of which only one is of spades.

Hence, if A be an event of getting an ace of spades.

n (A) = 1

So, P (A) = $\frac{n (A)}{n (S)} = \frac{1}{52}$ .

(c) (i) Let B be an event of drawing an ace. As there are 4 ace cards we have,

n (B) = 4

So, P (B) = $\frac{n (B)}{n (S)} = \frac{4}{52} = \frac{1}{13}$ .

(ii) Let D be an event of drawing black cards. Since there are 26 black cards we have,

n (D) = 26.

So, P (D) = $\frac{n (D)}{n (S)} = \frac{26}{52} = \frac{1}{2}$

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8. Is 3 ! + 4 ! = 7 ! ?

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8. L.H.S = 3! + 4!

= (1 * 2 * 3) + (1 * 2 * 3 * 4)

= 6 + 24

= 30

R.H.S = 7!

= 1 * 2 * 3 * 4 * 5 * 6 * 7

= 5040

As, L.H.S ≠ R.H.S

3! + 4! ≠ 7!

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Permutation and Combinations Ex 6.2 Solutions

7. Evaluate

(i) 8 !

(ii) 4 ! – 3!

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1. We know that, n! = n (n – 1) (n – 2)…….

i. 8!

8! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8

= 40320

ii. 4! – 3!

= (1 * 2 * 3 * 4) – (1 * 2 * 3)

= 24 – 6

= 18

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26. A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

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26. The sample space of throwing s dice is

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

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n (A)}{n (S)} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So, n (B) = 4

Therefore P (B) = $\frac{n (B)}{n (S)} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be event such that a number less than or equal to one will appear. Then,

C = {1}

So, n (C) = 1

? P (C) = $\frac{n (C)}{n (S)} = \frac{1}{6}$

(iv) Let D be event such that a number more than 6 appears. Then,

D =∅

So, n (D) = 0

? P (D) = $\frac{n (D)}{n (S)} = \frac{0}{6} = 0$

(v) Let E be event such that a number less than 6 appears. Then

E = {1, 2, 3, 4, 5}

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26. The sample space of throwing s dice is

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

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n (A)}{n (S)} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So, n (B) = 4

Therefore P (B) = $\frac{n (B)}{n (S)} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be event such that a number less than or equal to one will appear. Then,

C = {1}

So, n (C) = 1

? P (C) = $\frac{n (C)}{n (S)} = \frac{1}{6}$

(iv) Let D be event such that a number more than 6 appears. Then,

D =∅

So, n (D) = 0

? P (D) = $\frac{n (D)}{n (S)} = \frac{0}{6} = 0$

(v) Let E be event such that a number less than 6 appears. Then

E = {1, 2, 3, 4, 5}

So, n (E) = 5

? P (E) = $\frac{n (E)}{n (S} = \frac{5}{6}$

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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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