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A box contains 20 screws which are identical in size, but 12 of which are zinc coated and 8 of which are not. Two screws are selected at random, without replacement. a. Find the probability that both are zinc coated. b. Find the probability that at least one is zinc coated.

Short Answer

Expert verified
a. \( \frac{33}{95} \); b. \( \frac{81}{95} \)

Step by step solution

01

Determine Total Outcomes

We are picking two screws out of a total of 20 screws. Without replacement, the total number of ways to pick 2 screws from 20 is given by the combination formula: \[\binom{20}{2} = \frac{20 \times 19}{2} = 190\]
02

Determine Favorable Outcomes for Both Zinc Coated

For both screws to be zinc coated, first one selected is zinc-coated and the second one is chosen from the remaining zinc-coated screws. There are 12 zinc-coated screws initially:1. Probability of first being zinc-coated: \( \frac{12}{20} \).2. Probability of second being zinc-coated after one is removed: \( \frac{11}{19} \).Thus, the probability for both being zinc-coated is:\[\frac{12}{20} \times \frac{11}{19} = \frac{132}{380} = \frac{33}{95}\]
03

Calculate Probability of At Least One Zinc Coated

The probability of the event "at least one is zinc coated" is the complement of the event "none is zinc coated" (or both not zinc-coated):- Probability both not zinc-coated: - First is not zinc-coated: \( \frac{8}{20} \). - Second is not zinc-coated: \( \frac{7}{19} \).The product gives us the probability of both not zinc-coated:\[\frac{8}{20} \times \frac{7}{19} = \frac{56}{380} = \frac{14}{95}\]The probability of at least one being zinc-coated is the complement:\[1 - \frac{14}{95} = \frac{81}{95}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics that deals with counting the number of ways certain events can happen. It is essential for solving problems where we need to determine how many different ways we can pick items from a set. For example, when dealing with the scenario of selecting screws from a box, combinatorics enables us to know the total number of ways such events can occur.

In the given problem, we need to find out how many ways we can choose two screws out of the 20 available. This is where the combination formula is handy. The formula for a combination to choose two items out of a total of 20 items is given by:\[\binom{20}{2} = \frac{20 \times 19}{2}\]This calculates the result as 190 different possible ways to select two screws from the box. The combination formula helps us understand how quantity changes when the order of selection does not matter. This foundational tool is crucial in many probability problems.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already happened. It is an important concept in probability theory that allows us to recalculate the probability of an event when additional information is available.

In our exercise, the concept of conditional probability helps us determine the probability of drawing a second zinc-coated screw after one has already been removed from the box. Initially, there are 12 zinc-coated screws out of 20, so the probability of picking one is:\[\frac{12}{20}\]After selecting one zinc-coated screw, the situation changes: now there are 11 zinc-coated screws left out of a total of 19 screws:\[\frac{11}{19}\]This adjustment reflects conditional probability, as the probability of picking the second zinc screw depends on the first outcome—having already picked one zinc-coated screw. Conditional probabilities provide a more accurate picture by considering past events, which significantly influences outcomes.
Complementary Probability
Complementary probability is used to find the probability of the occurrence of the complementary event (i.e., the event not happening). Once we know the probability of an event occurring, we can find the probability of the complementary event simply by subtracting the given event's probability from 1.

In the exercise, we want to find the probability that at least one of the screws is zinc-coated. This involves understanding the complementary situation, where neither of the screws is zinc-coated (both screws are not zinc-coated).

The probability that both screws are not zinc-coated is calculated as follows:
  • Probability the first screw picked is not zinc-coated: \(\frac{8}{20}\)
  • Probability the second one is also not zinc-coated: \(\frac{7}{19}\)
The combined probability for this situation becomes:\[\frac{8}{20} \times \frac{7}{19} = \frac{14}{95}\]Complementary probability tells us that the chance of at least one of the screws being zinc-coated is:\[1 - \frac{14}{95} = \frac{81}{95}\]Using complementary probability allows us to solve complex problems by considering the easier-to-calculate complementary case.

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Most popular questions from this chapter

Individuals with a particular medical condition were classified according to the presence \((T)\) or absence \((N)\) of a potential toxin in their blood and the onset of the condition (E: early, M: midrange, L: late). The breakdown according to this classification is shown in the two-way contingency table. $$ \begin{array}{|c|c|c|c|} \hline & E & M & L \\ \hline T & 0.012 & 0.124 & 0.013 \\ \hline N & 0.170 & 0.638 & 0.043 \\ \hline \end{array} $$ One of these individuals is selected at random. Find the probability of each of the following events. a. The person experienced early onset of the condition. b. The onset of the condition was either midrange or late, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements. c. The toxin is present in the person's blood. d. The person experienced early onset of the condition and the toxin is present in the person's blood. e. The person experienced early onset of the condition or the toxin is present in the person's blood, in two ways: (i) by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.

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Make a statement in ordinary English that describes the complement of each event (do not simply insert the word "not"). a. In the roll of a die: "two or less." b. In the roll of a die: "one, three, or four." c. In two tosses of a coin: "at most one heads." d. In the random selection of a college student: "Neither a freshman nor a senior."

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