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Fair spinner A has 4 equal sections labeled 1,2,3,4. Fair spinner B has 4 equal labled 2,3,4,5. each spinner is spun once and the numbers are addedd. work out the probabilty that the total is 5 or less . give ur answer as a fraction.

Respuesta :

Answer:

P = 3/8

Step-by-step explanation:

Each spinner has 4 possible outputs, so if each spinner is spun once, we have a total of 4*4=16 outputs:

(1,2), (1,3), (1,4), (1,5),

(2,2), (2,3), (2,4), (2,5),

(3,2), (3,3), (3,4), (3,5),

(4,2), (4,3), (4,4), (4,5).

The total of these outputs is:

3, 4, 5, 6,

4, 5, 6, 7,

5, 6, 7, 8,

6, 7, 8, 9.

We have a total of 6 outputs where the total is 5 or less, so the probability of the total being 5 or less is:

P = N(5 or less) / N(total)

P = 6 / 16 = 3/8

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