An urn contains 4 red & 6 white chips, 5 chips are drawn at random without replacement. What is the probability that the 2nd chip is red? What's the probability of drawing a total of 2 red chips out of 5?
For the first question, “What is the probability that the 2nd chip is red?” there are two main approaches to solve this.
The first, and more intuitive (as well as lengthier), is simply looking at an outcome tree.
For the first draw, we have two possibilities: red or white.
Red has a 4/10 chance (4/(4+6)) and white has a 6/10 chance (6/(4+6), which reduce down to 2/5 and 3/5, respectively.
For the second draw, the denominator changes, and the numerator may change, based on the first draw.
If red was drawn on the first draw, the chance of drawing red on the second draw is (4–1)/(4+6–1) (we remove one red chip from both the number of red chips available and from the total chips available in the urn) which reduces down to 3/9 or 1/3.
If white was drawn on the first draw, the chance of drawing red on the second draw is 4/(4+6–1) (we remove one white chip from the total chips available in the urn) which reduces down to 4/9.
To get the probability that the 2nd chip is red from this, we add up the individual probabilities of arriving at a 2nd red chip.
If red was drawn on the first draw (2/5), and so the chance of drawing red on the second draw is 1/3, the chance of this path to a red on the second draw occurring is the product of these two probabilities (2/5 * 1/3) which comes to 2/15.
If white was drawn on the first draw (3/5), and so the chance of drawing red on the second draw is 4/9, the chance of this path to a red on the second draw occuring is the product of these two probabilities (3/5 * 4/9) which comes to 12/45, which reduces down to 4/15.
When we sum these two (2/15 + 4/15), we get 6/15, which reduces down to 2/5, or 40%.
The second method looks more directly at the expected outcome of the second draw. From going through the lengthier solution presented first, it becomes apparent that the odds of drawing a red on the second draw is the same as drawing a red on the first draw, and that’s a key hint to a more direct path.
Because there is a 2/5 (.4) chance of drawing red on the first draw, and a 3/5 (.6) chance of drawing white on the first draw, the expected urn composition on the second draw can be directly summarized as (4-.4 or 3.6) red chips and (6-.6 or 5.4) white chips, by removing the expected number of red chips and expected number of white chips, to give us the expected number remaining.
This gives us 3.6 red chips available from a pool of 9 (3.6+5.4), which comes to that same 2/5, or 40%.
By applying the lesson from the second solution of Part1, we can recognize that the order isn’t important here (all successful arrangements of outcomes will have the same chance of occurring).
For example, if we simplify Part2 to 2 draws instead of 5, and we want to know the odds of 1 red chip and 1 white chip being drawn
The odds of red then white are 4/10 * 6/9
The odds of white then red are 6/10 * 4/9
In both cases the numerator is 24 and the denominator is 90, because order doesn’t matter when calculating this product.
So the probability of 1 red and 1 white in the first two draws is simply the number of successful arrangements (2) multiplied by the chance of any one of the successful arrangements
Applying this to the actual Part2 question, we can calculate that there are 10 possible arrangements (4 with a red chip in position 1, 3 additional with a red chip in position 2 but not position 1, 2 additional with a red chip in position 3 but not positions 1 or 2, 1 additional with a red chip in position 4 but not positions 1, 2, or 3.
We also know that the numerators for a successful arrangement will be 6, 5, 4 (for the white draws) and 4, 3 (for the red draws), and the denominators will be 10, 9, 8, 7, 6, so the chance of any one successful arrangement is the product of (6,5,4,4,3) divided by the product of (10,9,8,7,6) which equals 1,440/30,240, which reduces to 1/21
Combining the pieces from b. and c., we get the odds of 2 red chips in 5 draws to be 10 * 1/21 which comes to 10/21, or approximately 47.62%
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00luvs11
Probability is the number of favorable outcomes divided by the number of total outcomes. The number of combinations is defined as (n!)/(r! * (n-r)!) where n is the total number of items in the set and where r is the number of items being chosen. There are three combinations we need to calculate: red, white, and total.
00luvs11
For red, we want 2 red out of a total possible of 4 red, so the number of favorable combinations is 4!/(2! * (4–2)!), which equals 24/4, or 6
00luvs11
For total, we want 5 items out of a total possible of 10 items, so the number of favorable combinations is 10!/(5! * (10–5)!), which equals 3,628,800/14,400, or 252 This means that our total probability of having 2 red chips and 3 white chips when drawing 5 chips is: (6*20)/252, which reduces to 10/21, or approximately 47.62%
00luvs11
yan lang alam ko hehe sorry kung mali basta mag tulongan kita
Answers & Comments
Answer:
An urn contains 4 red & 6 white chips, 5 chips are drawn at random without replacement. What is the probability that the 2nd chip is red? What's the probability of drawing a total of 2 red chips out of 5?
For the first question, “What is the probability that the 2nd chip is red?” there are two main approaches to solve this.
The first, and more intuitive (as well as lengthier), is simply looking at an outcome tree.
For the first draw, we have two possibilities: red or white.
Red has a 4/10 chance (4/(4+6)) and white has a 6/10 chance (6/(4+6), which reduce down to 2/5 and 3/5, respectively.
For the second draw, the denominator changes, and the numerator may change, based on the first draw.
If red was drawn on the first draw, the chance of drawing red on the second draw is (4–1)/(4+6–1) (we remove one red chip from both the number of red chips available and from the total chips available in the urn) which reduces down to 3/9 or 1/3.
If white was drawn on the first draw, the chance of drawing red on the second draw is 4/(4+6–1) (we remove one white chip from the total chips available in the urn) which reduces down to 4/9.
To get the probability that the 2nd chip is red from this, we add up the individual probabilities of arriving at a 2nd red chip.
If red was drawn on the first draw (2/5), and so the chance of drawing red on the second draw is 1/3, the chance of this path to a red on the second draw occurring is the product of these two probabilities (2/5 * 1/3) which comes to 2/15.
If white was drawn on the first draw (3/5), and so the chance of drawing red on the second draw is 4/9, the chance of this path to a red on the second draw occuring is the product of these two probabilities (3/5 * 4/9) which comes to 12/45, which reduces down to 4/15.
When we sum these two (2/15 + 4/15), we get 6/15, which reduces down to 2/5, or 40%.
The second method looks more directly at the expected outcome of the second draw. From going through the lengthier solution presented first, it becomes apparent that the odds of drawing a red on the second draw is the same as drawing a red on the first draw, and that’s a key hint to a more direct path.
Because there is a 2/5 (.4) chance of drawing red on the first draw, and a 3/5 (.6) chance of drawing white on the first draw, the expected urn composition on the second draw can be directly summarized as (4-.4 or 3.6) red chips and (6-.6 or 5.4) white chips, by removing the expected number of red chips and expected number of white chips, to give us the expected number remaining.
This gives us 3.6 red chips available from a pool of 9 (3.6+5.4), which comes to that same 2/5, or 40%.
By applying the lesson from the second solution of Part1, we can recognize that the order isn’t important here (all successful arrangements of outcomes will have the same chance of occurring).
For example, if we simplify Part2 to 2 draws instead of 5, and we want to know the odds of 1 red chip and 1 white chip being drawn
The odds of red then white are 4/10 * 6/9
The odds of white then red are 6/10 * 4/9
In both cases the numerator is 24 and the denominator is 90, because order doesn’t matter when calculating this product.
So the probability of 1 red and 1 white in the first two draws is simply the number of successful arrangements (2) multiplied by the chance of any one of the successful arrangements
Applying this to the actual Part2 question, we can calculate that there are 10 possible arrangements (4 with a red chip in position 1, 3 additional with a red chip in position 2 but not position 1, 2 additional with a red chip in position 3 but not positions 1 or 2, 1 additional with a red chip in position 4 but not positions 1, 2, or 3.
We also know that the numerators for a successful arrangement will be 6, 5, 4 (for the white draws) and 4, 3 (for the red draws), and the denominators will be 10, 9, 8, 7, 6, so the chance of any one successful arrangement is the product of (6,5,4,4,3) divided by the product of (10,9,8,7,6) which equals 1,440/30,240, which reduces to 1/21
Combining the pieces from b. and c., we get the odds of 2 red chips in 5 draws to be 10 * 1/21 which comes to 10/21, or approximately 47.62%
The number of combinations is defined as (n!)/(r! * (n-r)!) where n is the total number of items in the set and where r is the number of items being chosen.
There are three combinations we need to calculate: red, white, and total.
This means that our total probability of having 2 red chips and 3 white chips when drawing 5 chips is: (6*20)/252, which reduces to 10/21, or approximately 47.62%