Question

please solve, class 8, subject math, lesson data handling, explanation compulsory, no wrong answer, if you post wrong answer, then I will report​

Answer 1

Question:-

4) If two coins are tossed together, what is the probability of getting

a) both heads b) both tails c) one head and one tail

Given:-

• Two coins are tossed together.

Solution:-

When two coins are tossed together there is a probability of 4 outcomes. They are:-

• [Tail ,Tail] Or [T, T]
• [Tail, Head] Or [T, H]
• [Head, Head] Or [H, H]
• [Head, Tail] Or [H, T]

Now,

We know,

Probability = $\sf{\dfrac{Number\:of\:favourable\:Outcomes}{Total\:Number\:of\:Outcomes}}$

a) both heads

From all the possible outcomes, clearly see two heads together is only once.

Therefore,

P(of getting both heads) = $\sf{\dfrac{1}{4}}$

Therefore probability of getting both heads is $\sf{\dfrac{1}{4}}$

b) both tails

From all the possible outcomes, we can clearly see that two tails together is only once.

Therefore,

P(of getting both tails) =$\sf{\dfrac{1}{4}}$

Therefore, probability of getting both heads is $\sf{\dfrac{1}{4}}$

c) one head one tail.

From all the possible outcomes, we can clearly see that head and tail are together two times.

Therefore,

P(of getting one head one tail) = $\sf{\dfrac{2}{4} = \dfrac{1}{2}}$

Therefore, Probability of getting ine head one tail is $\sf{\dfrac{1}{2}}$

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Question:-

A dice is thrown once. What is the probability of getting:

a) a prime number b) a number between 2 and 6 c) an odd number d) an even number.

Given:-

• A dice is thrown once.

Solution:-

When a die is thrown, there is a probability of 6 outcomes. They are:-

1, 2, 3, 4, 5, 6

We know,

Probability = $\sf{\dfrac{Number\:of\:favourable\:Outcomes}{Total\:Number\:of\:Outcomes}}$

a) a prime number,

From all the favourable outcomes let us take out the prime numbers.

Prime numbers are:- 2, 3, 5

Therefore,

P(of getting a prime number) = $\sf{\dfrac{3}{6} =\dfrac{1}{2}}$

Therefore probability of getting a prime number is $\sf{\dfrac{1}{2}}$

b) a number between 2 and 6

From all the favourable outcomes, let us take out the numbers between 2 and 6

They are:- 3, 4, 5, 6

Therefore,

P(of getting a number between 2 and 6) = $\sf{\dfrac{4}{6} = \dfrac{2}{3}}$

Therefore probability of getting a number between 2 and 6 is $\sf{\dfrac{2}{3}}$

c) an odd number

From all the favourable outcomes let us take out the odd numbers.

Odd numbers:- 1, 3, 5

Therefore,

P(of getting an odd number) = $\sf{\dfrac{3}{6} = \dfrac{1}{2}}$

Therefore probability of getting an odd number is $\sf{\dfrac{1}{2}}$

d) an even number

From all the favourable outcomes, let us take out the even numbers.

Even numbers = 2, 4, 6

Therefore,

P(of getting an even number) = $\sf{\dfrac{3}{6} = \dfrac{1}{2}}$

Therefore, Probability of getting an even number is $\sf{\dfrac{1}{2}}$

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