Basic Probability supplementary material
Uniform Distribution
• Aboxof5redand5blueballs
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– Uniform distribution in color because all colors have the same percentage of balls.
• A box of 2 red, 2 blue, 2 green, 2 yellow, 2 white balls
– Uniform distribution in color because all colors have the same percentage of balls.
• A box of 2 red, 2 blue, 2 green, 4 white balls
– Not uniform distribution in color because all colors do not have the same percentage of balls.
• A box with unknown number of balls in 4 colors, where each color forms 25% of the total number of balls.
– Uniform distribution in color because all colors have the same percentage of balls.
• A box of 2 red, 2 blue, 2 green, 4 white balls. All red, blue, and 1 green ball weigh 50g.The remaining balls all weigh 100g.
– Not uniform distribution in color because all colors do not have the same percentage of balls.
– Uniform distribution in weight, because both weights (50g and 100g) have same percentage of balls.
Independent distributions
• Example 1: A box contains 50 pairs of boots, 20 of them red, 30 of them black.
– Color of the boot and the leg (left or right) are distributions independent of each other.This is because if you take all the red color boots, half of them (50%) will be left and other half will be right. The same applies for the black color books.
– Similarly, if you pick all the left leg boots, 20 of them (40%) will be red and 30 of them (60%) will be black. Similarly, if you pick the right leg boots, 20 of them will be red and 30 of them will be black.
– Since in both the above scenarios, the proportions/percentages remain the same we can say that the color and leg distributions are independent of each other.
Dependent distributions
• Example 2: A box contains 50 pairs of boots, 20 of them red, 30 of them black. But 10 of the red pairs are left for both the boots (bad shoe maker).
– Now if you take only the red colored shoes, there are 30 left boots (75%) and 10 right boots (25%). But if you take the black colored boots, there are 30 left boots (50%) and 30 right boots (50%). Clearly the percentages do not align. Hence in this case color and leg distributions are not independent of each other.
Working with independent distributions
• In Example 1, what is the percentage of boots that are both red and left ?
– We know that half of all the boots are left(0.5), we also know that 20/(20+30) = 0.4 fraction of boots are red. Since the distributions are independent of each other, we will have 0.5 x 0.4 = 0.2 or 20% of the boots will be both red and left.
– Not convinced ?
• Number of red boot pairs = 20. – 20left,20rightboots.
• Number of black boot pairs = 30. – 30left,30rightboots
• So what is the percentage of left red boots ?
– = number of left red boots / total number of boots. – =20/(20+20+30+30)=0.20,i.e20%
Working with dependent distributions
• In Example 2, what is the percentage of boots that are both red and left ?
– Weknowthatnumberofredbootpairsis20,whichis0.4fractionofthetotal number of boots [ 20 / (20 + 30 ) ].We are also told that 10 of the red pairs are left on both boots due to manufacturing defect.This means we have 30 left boots and 10 right boots for red, i.e. a fraction of 30/(30+10) = 0.75 ( 75%) of red boots are left.Therefore the total percentage of red left boots in this case will be 0.4 x 0.75 = 0.3 or 30%
– Notconvinced?
• Number of red boot pairs = 20 – 30 left, 10 right
• Number of black boot pairs = 30 – 30 left, 30 right.
• Percentage of left red boots ? Is same equation as before – = number of left red boots / total number of boots.
– 30/(30+10+30+30)=0.30or30%
– Noticehowinthisexampleweignoredtheoveralldistributionofleftboots but worked with the information of what percentage of red boots are left ? That is because this percentage was dependent on what color the boot is.
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