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Eva8 [605]
3 years ago
5

The students at a school assembly

Mathematics
1 answer:
jekas [21]3 years ago
4 0

Answer:

292

Step-by-step explanation:

73 students in 1 group

4 groups all together

73 Students times 4 groups = amount of students all together

so 73x4=292

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Suppose you and a friend each choose at random an integer between 1 and 8, inclusive. For example, some possibilities are (3,7),
Bezzdna [24]

Answer and explanation:

Given : Suppose you and a friend each choose at random an integer between 1 and 8, inclusive.

The sample space is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)  (1,7) (1,8)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)   (2,7) (2,8)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)   (3,7) (3,8)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)   (4,7) (4,8)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)   (5,7) (5,8)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)   (6,7) (6,8)

(7,1) (7,2) (7,3) (7,4) (7,5) (7,6)   (7,7) (7,8)

(8,1) (8,2) (8,3) (8,4) (8,5) (8,6)   (8,7) (8,8)

Total number of outcome = 64

To find : The following probabilities ?

Solution :

The probability is given by,

\text{Probability}=\frac{\text{Favorable outcome }}{\text{Total outcome}}

a) p(you pick 5 and your friend picks 8)

The favorable outcome is (5,8)= 1

\text{Probability}=\frac{1}{64}

b) p(sum of the two numbers picked is < 4)

The favorable outcome is (1,1), (1,2), (2,1)= 3

\text{Probability}=\frac{3}{64}

c) p(both numbers match)

The favorable outcome is (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8) = 8

\text{Probability}=\frac{8}{64}

\text{Probability}=\frac{1}{8}

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