AboutJosh Expertise When I work through problems, I emphasize principles and key ideas which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponentials & Logarithms. Basic Calculus. Probability, Permutations and Combinations. Mathematical Induction. Complex numbers. Physics problems.
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Experience: I have worked as a teaching assistant in college. My hope is that more people will share knowledge without boundary, give help without seeking recognition or monetary rewards.
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Education Credentials: Bachelor degree in Engineering Science."Everyone struggles with something."
Question I have 3 questions I can't seem to work out.
1. 2 cards are simultaneously drawn from an ordinary deck. The Probability that there're NOT both the same suit is??
2. A test shows that 10 out of 100 gadgets are defective. In a random sample of two gadggets are chosen, the probability that both are defective it is:
3. A drawer contains red socks, black socks, and white socks. What is the least number of socks that must be taken out of the drawer to be sure of having 7 matching pairs of socks?
If you could break them down for me that would be wonderful. Thanks :)
Answer Hi Meredith,
Given the limited space we have, I will start with the assumption that you are already quite familiar with permutation and combination. If this is not the case, you can always do a search on Google later, if and when necessary. There are plenty of examples and tutorials on the web which explain these concepts.
For Q1, the first card we draw must belong to one of four suits. It doesn't matter if it is a spade, heart, club or diamond. For illustration purpose, let's suppose this card is a "spade". Then, we are left with 12 cards of "spade" and 13 cards each of "heart", "club" and "diamond". The probability consideration only begins when we draw the second card. In this example, there is a (3*13) out of (3*13+12) chance that the second card is NOT a "spade". The situation is identical regardless of the suit of the first card. So, the probability is 39/51. This question falls into the category of "selection without replacement" and the population size is finite.
For Q2, we first compute the probability that the item is "defective". Clearly, we can set p(d)=0.1 (from 10/100). It is important to note that we are dealing with a statistic here, viz., the average failure rate. This is different from the counting argument used when we pick something from a small batch of fixed size. If we pick items from a sample of fixed size (like in question 1), the size of the sample shrinks as cards are picked without replacement. The keyword here is "random" and presumably we have a VERY large population space. Removing an item (sampling) from this population does not change the statistic. Finally, choosing two items independently, the joint probability that both are defective is given by the product of the average failure rate for an individual item selected from the population. i.e., P("d,d")=p(d)*p(d)=0.01.
Q3 is more difficult to answer without knowing the exact color composition in a small population. One might approach this using trinomial expansion [http://en.wikipedia.org/wiki/Trinomial_expansion] if the numbers are known. If by "sure", you mean with a (1-alpha)% level of confidence that "matching is achieved", then a probabilistic argument may be given using the formula. Like in Q1, the sample size will continue to shrink as we pull socks from the drawer. Due to finite effect, we may have to consider conditional dependence in the probability calculations as well. Here is an example to clarify this.
Suppose there are 41 pairs of socks, of which 20 are "black", 20 are "white" and only a single pair is "red". Let's say that we have, by luck, already selected 6 matching pairs (3 "black" and 3 "white"). Then, we are still left with 1 pair of "red" amongst 17 pairs of "black" and 17 pairs of "white". Intuitively, we have a long way to go until we hit the "red" pair. On the other hand, if we were fortunate enough to have drawn the "red" pair early (this happens on average with a small probability), having picked a pair of "red", 3 pairs of "black" and 2 pairs of "white" already, then, it won't be long to go before we reach another "white" pair. As you can see, the relative composition of color, and conditional dependencies both impact on the calculation.
On the other hand, if by "sure" you mean absolute certainty, in the worse case scenario, we may have to almost empty the drawer if we are unlucky (and hit the "red" pair last as considered in our example). And nothing short of this is going to guarantee 100% the requirements will be met. There is a lot of "if's" and "but". It is unclear whether the 7 matching pairs can be of any color assortment or must cover at least one of each color. Perhaps the question is really intended to canvas ideas and be answered with a bit of lateral thinking.
