probability

When arranging 3 nos. out of 6 different nos., no. of possible permutations = 6P3 = 120

a) To be a multiple of 10, the last digit must be 0 and then the remaining 2 digits can be arranged randomly from the remaining 5 available nos.

So no. of possible permutations = 5P2 = 20

Prob = 20/120 = 1/6

b) The first digit must be 4, 5 or 9, so no. of possible permutations = 3 x 5P2 = 60

Prob = 1/2

c) The first digit can be 4, 5 or 9 and then the last digit must be 0

No. of possible permutations = 3 x 4 x 1 = 12 (since when the first and last digits are fixed, the middle digit has 4 possibilities)

Probability = 1/10

a)P(the number is a multiple of 10)

=P( the first card drawn is any number other than 0, the second card drawn is any number other than 0, the last card drawn must be 0)

=(5/6)(4/5)(1/4)
=1/6

b) P(the number is greater than 400)
=P(the first card drawn is greater than or equal to 4) = (3/6) =1/2

c)P(the number is a multiple of 10 and greater than 400)
=P(the first card drawn is greater than or equal to 4, the last card drawn must be 0)

=(3/6)(1/5)
=1/10

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