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
