Which experiment has equally likely outcomes?
A.Choose a number at random from 1 to 7
B.Toss a coin
C.Roll a die
D.All the above

Hint: When the experiment has the same chance of happening i.e., the probability of each event is equalis known as equally likely event.

Complete step-by-step answer:
First of all, let us discuss what are equally likely events.
When all the events happening give the same output and their probability is also the same. These types of events are called equally likely events or experiments have equally likely events.
Example: - choosing a heart and choosing a spade from a deck of cards are the equally likely events. This experiment has equally likely outcomes.

Formula for probability is given as follow:
$ \Rightarrow \ Probability = \dfrac{{FavourableOutcomes}}{{TotalOutcomes}}$
Let's start with the option A, we have to choose a number at random from 1 to 7. Starting with 1. Here total outcomes are 7 i.e. 1,2,3,4,5,6 and 7 and favourable outcome is 1 for each number. So, probability of each number is
$ \Rightarrow \operatorname{P} (1) = \dfrac{1}{7}$
$ \Rightarrow \operatorname{P} (2) = \dfrac{1}{7}$
$ \Rightarrow \operatorname{P} (3) = \dfrac{1}{7}$
$ \Rightarrow \operatorname{P} (4) = \dfrac{1}{7}$
$ \Rightarrow \operatorname{P} (5) = \dfrac{1}{7}$
$ \Rightarrow \operatorname{P} (6) = \dfrac{1}{7}$
$ \Rightarrow \operatorname{P} (7) = \dfrac{1}{7}$
You can see the probabilities for all are the same so there are equally likely outcomes. So, this option is right.
Now take option B, we are tossing a coin. Here total outcomes are 2 i.e. Head and Tail and favourable outcome is 1 for coming only head and only tail. So, probability of each is
$ \Rightarrow \operatorname{P} (H) = \dfrac{1}{2}$
$ \Rightarrow \operatorname{P} (T) = \dfrac{1}{2}$
You can see the probabilities both are the same so there are equally likely outcomes. So, this option is also right.
Now take option B, we are rolling a dice. Here total outcomes are 6 i.e. 1, 2, 3, 4, 5 and 6 and favourable outcome is 1 for coming only 1, 2, 3, 4, 5 and 6. So, probability of each is
$\begin{gathered}
   \Rightarrow \operatorname{P} (1) = \dfrac{1}{6} \\
   \Rightarrow \operatorname{P} (2) = \dfrac{1}{6} \\
   \Rightarrow \operatorname{P} (3) = \dfrac{1}{6} \\
   \Rightarrow \operatorname{P} (4) = \dfrac{1}{6} \\
   \Rightarrow \operatorname{P} (5) = \dfrac{1}{6} \\
   \Rightarrow \operatorname{P} (6) = \dfrac{1}{6} \\
\end{gathered} $
You can see the probabilities all are the same so there are equally likely outcomes. So, this option is also right.
Here, you can see that all the options A, B and C are right. Option D-all of the above is right.
So, the correct option is D.

Note: While solving this question you can directly judge that in first there are 7 numbers and all numbers are different and favourable outcomes for each number is the same. So, they are equally likely events. In the second head can occur 1 time out of 2 and tail can occur 1 time out of 2 i.e. the chance of both are 50%. So, they both are equally likely events. Similarly, in the case of death they are equally likely. Common mistakes are choosing only option A as students generally go for a single option. But if one of the options is all of these, then we have to see each option before conclusion.