Question

Enter an inline fraction (a/b) in simplest form. A box contains identical labels that each have one character written on it.

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Answer #1

There are total 8 cards, out of them E is twice.

1

If the labels are drawn without replacement :

Probability of drawing G in first = 1/8

Probability of drawing R in second = 1/7

(since in second draw, one label is already taken out)

Probability of drawing E in third = 2/6

(since there are two labels with E character)

Probability of drawing E in fourth = 1/5

(since one label with E is taken out)

Probability of drawing N in fifth = 1/4

Probability of drawing S in sixth = 1/3

Hence probability that labels are drawn in order GREENS

= (1/8)*(1/7)*(2/6)*(1/5)*(1/4)*(1/3)

= 2/20160

= 1/10080

2.

If labels are drawn with replacement :

Probability of drawing G in first = 1/8

Probability of drawing R in second = 1/8

(since before second draw, label with G is kept again )

Probability of drawing E in third = 2/8

(since there are two labels with E character)

Probability of drawing E in fourth = 2/8

(since one label with E is kept again in box before fourth draw)

Probability of drawing N in fifth = 1/8

Probability of drawing S in sixth = 1/8

Hence probability that labels are drawn in order GREENS

= (1/8)*(1/8)*(2/8)*(2/8)*(1/8)*(1/8)

= 4/ 262144

= 1/65536

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