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Question

Answers

$

(a){\text{ }}\dfrac{1}{3} \\

(a){\text{ }}\dfrac{1}{4} \\

(a){\text{ }}\dfrac{1}{2} \\

(a){\text{ }}\dfrac{2}{3} \\

$

Answer
Verified

Hint- In this problem we are rolling up a dice and tossing a coin simultaneously. We have to find the probability of getting a head and an odd number, so write all the possible sample cases that the event of rolling a dice and throwing up a coin can have. Amongst these all possible events just take out the one with an odd number and head. Then use the basic probability formula to reach the answer.

Complete step-by-step answer:

Let ${\text{E}}$ be the event of getting a head and an odd number after rolling up a dice and throwing up a coin simultaneously.

So total possible sample points for event E will be (1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T), (5, H), (5, T), (6, H), (6, T).

Thus ${\text{n(S) = 12}}$……………………… (1)

Now the favorable cases for event E will be the one having head and an odd number, so they are (1, H), (3, H), (5, H).

Thus ${\text{n(E) = 3}}$…………………….. (2)

Now using the basic formula of probability that probability of an event A is ${\text{P(A) = }}\dfrac{{{\text{favorable outcomes}}}}{{{\text{total number of outcomes}}}} = \dfrac{{{\text{n(A)}}}}{{{\text{n(S)}}}}$………………………… (3)

So probability of event E will be${\text{P(E)}}$, using equation (1)

${\text{P}}\left( {\text{E}} \right) = \dfrac{{{\text{n(E)}}}}{{{\text{n(S)}}}}$………………… (4)

On substituting the values from equation (1) and (2) we get,

${\text{P}}\left( {\text{E}} \right) = \dfrac{3}{{12}} = \dfrac{1}{4}$

So the probability of getting a head and an odd number after rolling up a dice and throwing up a coin simultaneously is $\dfrac{1}{4}$.

Thus option (b) is the right answer.

Note – Whenever we face such types of problems the key concept is to have the understanding of the basic probability formula. This along with the all possible sample cases with the favorable cases as per the question requirement will help us to get the answer.

Complete step-by-step answer:

Let ${\text{E}}$ be the event of getting a head and an odd number after rolling up a dice and throwing up a coin simultaneously.

So total possible sample points for event E will be (1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T), (5, H), (5, T), (6, H), (6, T).

Thus ${\text{n(S) = 12}}$……………………… (1)

Now the favorable cases for event E will be the one having head and an odd number, so they are (1, H), (3, H), (5, H).

Thus ${\text{n(E) = 3}}$…………………….. (2)

Now using the basic formula of probability that probability of an event A is ${\text{P(A) = }}\dfrac{{{\text{favorable outcomes}}}}{{{\text{total number of outcomes}}}} = \dfrac{{{\text{n(A)}}}}{{{\text{n(S)}}}}$………………………… (3)

So probability of event E will be${\text{P(E)}}$, using equation (1)

${\text{P}}\left( {\text{E}} \right) = \dfrac{{{\text{n(E)}}}}{{{\text{n(S)}}}}$………………… (4)

On substituting the values from equation (1) and (2) we get,

${\text{P}}\left( {\text{E}} \right) = \dfrac{3}{{12}} = \dfrac{1}{4}$

So the probability of getting a head and an odd number after rolling up a dice and throwing up a coin simultaneously is $\dfrac{1}{4}$.

Thus option (b) is the right answer.

Note – Whenever we face such types of problems the key concept is to have the understanding of the basic probability formula. This along with the all possible sample cases with the favorable cases as per the question requirement will help us to get the answer.

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