Question

1 If a coin is tossed three times and the total number of tails are observed, what is the sample space for the experiment?.

2 Relative frequency approach is the same as……..

3 A bag contains 2 natural oranges and 3 artificial oranges. Each of four persons, A, B, C, and D, in the order named, draws one orange and does not replace it. The first to draw a natural orange receives N20. What is the probability that A, B, and C lose and D wins

4 If a die is thrown once, what is the probability of getting an even number?.

5. 4 different Biology books, 6 different CIT books and 2 different Mathematics books are to be arranged on a shelf. How many different arrangements are possible if only the Biology books must stand together?

Answer 1

Answer on Question #59916 – Math – Statistics and Probability

Question

1. If a coin is tossed three times and the total number of tails are observed, what is the sample space for the experiment?

Answer:

$\{(T,T,T),(T,T,H),(T,H,H),(T,H,T),(H,T,T),(H,T,H),(H,H,T),(H,H,H)\}$.

Question

2. Relative frequency approach is the same as...

Answer: Relative frequency approach is the same as probability.

Question

3. A bag contains 2 natural oranges and 3 artificial oranges. Each of four persons, A, B, C, and D, in the order named, draws one orange and does not replace it. The first to draw a natural orange receives N20. What is the probability that A, B, and C lose and D wins.

Answer: $P = \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{10} = 0.1$.

Question

4. If a die is thrown once, what is the probability of getting an even number?

Answer: $P = \frac{3}{6} = \frac{1}{2} = 0.5$.

Question

5. 4 different Biology books, 6 different CIT books and 2 different Mathematics books are to be arranged on a shelf. How many different arrangements are possible if only the Biology books must stand together?

Solution

First arrange the Biology books. There are 4! ways to do this. Since they have to stay together - consider the Biology subject now as 1 big "book", 6 CIT books and 2 Math books.

Then we have $1 + 6 + 2 = 9$ "books" and we can arrange them in 9! ways. So we shall have $4!9! = 8,709,120$ arrangements.

Answer: 4!9!=8,709,120.

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