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ololo11 [35]
2 years ago
9

Arun has 25 chocolates with him. He had given 19 to his friends. What fraction of chocolates are with him?

Mathematics
2 answers:
lukranit [14]2 years ago
6 0
This is very easy, he has he has a remainder 6 chocolates
babunello [35]2 years ago
5 0
25-19=6
6/25 of the chocolates
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Answer:

<u><em></em></u>

  • <u><em>Event A: 1/35</em></u>
  • <u><em>Event B: 1/840</em></u>

<u><em></em></u>

Explanation:

<u>Event A</u>

For the event A, the order of the first 4 acts does not matter.

The number of different four acts taken from a set of seven acts, when the order does not matter, is calculated using the concept of combinations.

Thus, the number of ways that the first <em>four acts</em> can be scheduled is:

          C(m,n)=\dfrac{m!}{n!(m-n)!}

         C(7,4)=\dfrac{7!}{4!(7-4)!}=\dfrac{7!}{4!(3)!}=35

And<em> the number of ways that four acts is the singer, the juggler, the guitarist, and the violinist, in any order</em>, is 1: C(4,4).

Therefore the<em> probability of Event A</em> is:

           P(A)=1/35

Event B

Now the order matters. The difference between combinations and permutations is ordering. When the order matters you need to use permutations.

The number of ways in which <em>four acts </em>can be scheculed when the order matters is:

           P(m,n)=\dfrac{m!}{(m-n)!}

         P(m,n)=\dfrac{7!}{(7-4)!}=P(m,n)=\dfrac{7!}{4!}=840

The number of ways <em>the comedian is first, the guitarist is second, the dancer is third, and the juggler is fourth</em> is 1: P(4,4)

Therefore, <em>the probability of Event B</em> is:

            P(B)=1/840

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