12) There are 3 white balls, 4 green balls, 5 red balls, and 6 blue balls in a bag. If 2 balls are drawn, with replacement being made after the first ball is drawn. Find 12a) P(red and blue) 12b) P(both are red) 12a) I calculated 5/18 * 6/18 + 6/18 * 5/18=(5/18 * 6/18) *2 = 0.1852 on a test and got 3 marks deduced off for this question. I am very frustrated. Please Register or Log in to view the hidden image! I was talking to my instructor about that but he claims that the probability shouldn't be multiplied by 2 and asks me why didn't I multiply the question to 12b by 2 also. One of his arguments is that most students did not multiply it by 2. :bugeye: For 12b) I did 5/18 * 5/18=0.07716, do I actually have to multiply this by 2 if ORDER matters? Is there a powerful alternate solution (another way) of doing question 12a) ? What solid arguments can I provide to support my answer and to convince my instructor? Does the probability 5/18 * 6/18 include the cases in which the blue ball is drawn first and then a red ball is drawn? My interpretation of question 12a) is that as soon as you have a red ball and a blue ball after 2 draws, you are fine, these are the favorable outcomes. I strongly believe that my original answer to 12a) is correct...I will have to "debate" with my instructor tomorrow. Could someone help me out? Thank you! Please Register or Log in to view the hidden image!
You can argue that the exam question was ambiguous (again Please Register or Log in to view the hidden image! ). Is order important or not? Your answer is correct if order is not important. In general, if order is not important, you multiply the probability of getting all the colours in one particular order by the number of possible orders (ie. there are 2 ways of drawing a red and a blue ball: RB and BR, 6 ways of drawing 3 different coloured balls, etc.) There's only one way of getting 2 reds though, so there's no reason to double your final answer. (NB: this only works if the balls are replaced after each draw)
12) But for probability questions, order should always be important (i.e. permutations), right? Consider a bag of 1 red ball and 1 blue ball. 2 balls are drawn with replacement. The probability of drawing a red and a blue ball is (1/2 * 1/2) * 2= 2/4 (RB, BR) The probability of drawing 2 red balls is 1/2 * 1/2 = 1/4 (RR) Listing all the possible outcomes confirm this nicely: RR, RB, BR, BB If order is not important (i.e. combinations), the possible outcomes become: {R,R}, {R,B}, {B,B}, but we know that the probability of drawing a red ball and a blue ball is NOT 1/3.
The probability that the first ball drawn is red and that the second ball is blue (order is important) is 1/4. The probability that one red and one blue ball is drawn (order is not important) is 1/2. P(red and blue) could be interpreted either way.
distribution probablity any help Roll two dice and denote X the higher number and Y the lower number shown. (a) Is the joint distribution of X; Y discrete, continuous or neither? Explain. (b) Find IP(X = Y ), IP(X > 2Y ).
Calculate the probability that a randomly-selected person currently suffers side-effects: $(1/2 \times 1/400) + (1/2 \times 1/1200) = 1/600$ If Acetylin is taken off the market the probability the same person suffers side-effects becomes: $1 \times 1/1200 = 1/1200$ which is half as great as before. Phew! I'm glad we got that solved! It's been a long eleven-and-something year wait.
Here is a probability teaser I know the answer and how to work out and have tested it However am interested in those who might like to guess the odds and hope those fluent in maths don't spoil it for those who guess I will say YES or NO to those who guess or even those who work out a answer OFF POST and only post the result Also don't obtain items needed and run a simulation as almost same as pencil paper work out OK Two packs normal cards less the blanks and jokers Each random shuffled Placed alongside each other, two piles face down Each card on top of each pile is turned over at same moment Continue until all cards are face up, that is counted as one go Guess, or work out off post, odds of turning up 2 matching cards Matching means IDENTICAL, not any suite 4 to any other suite 4 If one pack you turn up 4 spades other pack has to turn up 4 spades So looking for odds any card turned up will have to its counterpart in other pile turn up at same moment GO Please Register or Log in to view the hidden image!
A couple button pushes on a calculator, no simulation needed. I got 63.57% odds that at least one pair of cards matches.
Pushing buttons on calculator is a simulation Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
You and I have a different definition of a simulation. I simply evaluated an arithmetic expression that could be typed in 11 characters, but is beyond my ability to do in my head.
This is an interesting puzzle that can be solved binarily (it's a word). Conditions: A: The first is a girl B: the next 3 are boys 0=g Possible outcomes and matches: 0: 0000 A 1: 0001 A 2: 0010 A 3: 0011 A 4: 0100 A 5: 0101 A 6: 0110 A 7: 0111 A B 8: 1000 9: 1001 10: 1010 11: 1011 12: 1100 13: 1101 14: 1110 15: 1111 B The first eight meet condition A. Of those, #7 also meets condition B. As does #15. Google: "When two events, A and B, are not mutually exclusive, the probability that A or B will occur is the sum of the probability of each event, minus the probability of the overlap." https://www.mathgoodies.com/lessons/vol6/addition_rules In other words, the probability is simply the sum of outcomes that meet any of the conditions (the sum automatically eliminates overlap). 9 out of 16 meet at least one condition, thus the probability is 9/16 or 0.5625.
