Question

Explaination needed

Q.1. Compute the probability of the occurrence of an event if the probability the event not occurring is 0.56.

Q.2. In a factory of 364 workers, 91 are married. Find the probability of selecting a worker who is not married.

Q. 3. From a deck of cards, 10 cards are picked at random and shuffled. The cards are as follows:

6, 5, 3, 9, 7, 6, 4, 2, 8, 2

Find the probability of picking a card having value more than 5 and find the probability of picking a card with an even number on it.

Q.4. From a bag of red and blue balls, the probability of picking a red ball is x/2. Find “x” if the probability of picking a blue ball is ⅔.

Q.5. Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and the number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. What is the probability of getting ‘Two tails’.

Class 9​

Answer 1

Answer:

Q.1. Compute the probability of the occurrence of an event if the probability the event not occurring is 0.56.

Solution:

Given,

P(not E) = 0.56

We know that,

P(E) + P(not E) = 1

So, P(E) = 1 – P(not E)

P(E) = 1 – 0.56

Or, P(E) = 0.44

Q.2. In a factory of 364 workers, 91 are married. Find the probability of selecting a worker who is not married.

Solution:

Given,

Total workers (i.e. Sample space) = n(S) = 364

Total married workers = 91

Now, total workers who are not married = n(E) = 364 – 91 = 273

Method 1: So, P(not married) = n(E)/n(S) = 273/364 = 0.75

Method 2: P(married) + P(not married) = 1

Here, P(married) = 91/364 = 0.25

So, 0.25 + P(not married) = 1

P(not married) = 1 – 0.25 = 0.75

Q. 3. From a deck of cards, 10 cards are picked at random and shuffled. The cards are as follows:

6, 5, 3, 9, 7, 6, 4, 2, 8, 2

Find the probability of picking a card having value more than 5 and find the probability of picking a card with an even number on it.

Solution:

Total number of cards = 10

Total cards having value more than 5 = 5

i.e. {6, 9, 7, 6, 8}

Total cards having an even number = 6

i.e. {6, 6, 4, 2, 8, 2}

So, the probability of picking a card having value more than 5 = 5/10 = 0.5

And, the probability of picking a card with an even number on it = 6/10 = 0.6

Q.4. From a bag of red and blue balls, the probability of picking a red ball is x/2. Find “x” if the probability of picking a blue ball is ⅔.

Solution:

Here, there are only red and blue balls.

P(picking a red ball) + P(picking a blue ball) = 1

x/2 + ⅔ = 1

=> 3x + 4 = 6

=> 3x = 2

Or, x = ⅔

Q.5. Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and the number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. What is the probability of getting ‘Two tails’.

Solution:

Given,

Total number of outcomes = Sample space = 360

Now, assume that the number of times ‘No Tail’ appeared to be “x”

So, the number of times ‘2 Tails’ appeared = 3x (from the question)

Also, the number of times ‘1 Tail’ appeared =2x (from the question)

As the total outcomes = 360,

x + 2x + 3x = 360

=> 6x = 360

Or, x = 60

∴ P(getting two tails) = (3 × 60)/360 = ½

Answer 2

Answer:

Step-by-step explanation:

1. If the event not occurring is 0.56, we can say that it is actually 56/100. So, total probability is 100, we can do this as subtracting 56 from 100. 100-56 = 44. so 44/100 = 0.44 is the required answer.
2. Workers which are not married, will be those which are not given out of 364 workers. So, it will be 364 - 91 = 273, so the probability will be 273/364, = 0.75.
3. Value more than 5 is 6, 5, 3, 9, 7, 6, 4, 2, 8, 2 ie 5. Total cases are 10. So, the probability will be the given outcome by the total outcome $\sf{\dfrac{Probable \ Outcome}{Total \ Outcome}}$ = 5/10 which is equal to 1/2 ie 0.5
4. Refer to attachment, whole the whole explanation.
5. Total outcomes = 360

Out of which, the given cases. Let us assume the number of times 2 Tails appeared as x. So, we can say that:-

3x = No tail

2(3x) = 1 Tail

Surely, if we combine 2 Tails, 1 Tail and No Tail, we will get 360. Now,

$\tt{\dfrac{3x + x + 6x}{360} = 1}$

$\tt{\dfrac{10x}{360} = 1}$

$\tt{10x = 360}$

$\tt{ x= 36 \ is \ the \ required \ answer.}$