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Maurinko [17]
2 years ago
3

Here are seven tiles Tom takes a tile at random. He does not replace the tile. Tom then takes at random a second tile. a) Calcul

ate the probability that both tiles Tom takes have the number 1 on them. b) Calculate the probability that the number on the second tile Tom takes is greater than the number on the first tile he takes.
Mathematics
1 answer:
wlad13 [49]2 years ago
6 0

Answer:

a) 1/21

b) 8/21

Complete question:

There are seven tiles:

1,1,3,3,3,5,5

Tom takes a tile at random. He does NOT replace the tile.

Tom then takes another tile at random.

a) Calculate the probability that both tiles Tom takes have the number 1 on them. b) Calculate the probability that the number on the second tile Tom takes is greater than the number on the first tile he takes.

Step-by-step explanation:

Total number of tiles = 7

Let Probability of having number 1 on the tiles = Pr (having 1)

Pr (having 1) = (number of times 1 appears on tiles)/(total number of tiles)

Number of times 1 appears on tiles = 2

Pr (having 1) = 2/7

Two tiles are drawn without replacement:

Probability of both tiles having number 1 on them = Pr (having 1 for both 1st and 2nd time)

= Pr (having 1) × Pr (having 1)

Since it is without replacement, the numbers in the second pick would reduce by 1 in both the numerator and denominator since we are picking same number. That is from 7 to 6 and from 2 to 1 to reflect that it was replaced.

= 2/7 × 1/6

= 2/42

Probability of both tiles having number 1 on them = 1/21

b) If 1st tile = 1, the second tile could be = 3 or 5

The pairs = Pr(1 and 3) and Pr(1 and 5)

Where Pr = probability

The probability is still without replacement. For both probability, the numbers in the second pick would reduce by 1 in the denominator since we are picking different numbers. That is from 7 to 6

Number of times 3 appears on tiles = 3

Number of times 5 appears on tiles = 2

Pr(1 and 3) = (2/7 × 3/6) = 1/7

Pr(1 and 5) = (2/7 × 2/6) = 2/21

If 1st tile = 3, the second tile = 5

Pr(3 and 5) = (3/7 × 2/6) = 1/7

If 1st tile = 5, the second tile = 0 (no number is greater than 5

Pr(5 and 0) = 0

Probability that the number on the second tile Tom takes is greater than the number on the first tile he takes = Pr(1 and 3) + Pr(1 and 5) + Pr(3 and 5) + Pr(5 and 0)

= 1/7 + 2/21 + 1/7 + 0

= (3+2+3)/21 = 8/21

Probability that the number on the second tile Tom takes is greater than the number on the first tile he takes = 8/21

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