Word problems are a nightmare HELP?

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SUMMARY

The discussion focuses on solving three distinct mathematical problems involving bacteria growth, consecutive integers, and probability with coin tosses. The first problem reveals that the initial number of bacteria was 14, calculated using the formula 275,562 = x * 3^9. The second problem requires algebraic formulation and quadratic solving to find five consecutive integers. The third problem explores the probability of getting heads when tossing three quarters, concluding that the probability is 6/7, given at least one tail appears. The conversation emphasizes the importance of understanding conditional versus joint probabilities.

PREREQUISITES
  • Understanding of exponential growth, specifically in biological contexts.
  • Basic algebra and quadratic equations for solving integer problems.
  • Fundamentals of probability theory, particularly conditional probability.
  • Familiarity with combinatorial analysis in probability scenarios.
NEXT STEPS
  • Study exponential growth models in biology, focusing on applications like bacteria population dynamics.
  • Learn how to solve quadratic equations and explore their applications in number theory.
  • Deepen knowledge of probability theory, specifically conditional probability and its implications.
  • Examine combinatorial methods in probability, including permutations and combinations in various scenarios.
USEFUL FOR

Students, educators, and professionals in mathematics, particularly those focusing on algebra, probability, and biological modeling. This discussion is beneficial for anyone looking to enhance their problem-solving skills in these areas.

vadlamudit
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hi

1.) The number of bacteria in a jar triples every twenty seconds. After three minutes, 275,562 bacteria are in the jar. How many bacteria were in the jar at the beginning of the experiment?

2.) Of a set of five consecutive integers, the sum of the squares of three smallest equals the sum of the squares of the two largest. Find all possible values for the five integers.

3.) If three quarters are tossed and if at least one lands tails up, what is the probability that at least one lands heads up?

pplese help. or i will be !1
 
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1. basically, we have 275562 = x * 3^9
therefore, x = 14, so there were 14 bacteria in the first place.

2. write this algebraically. x^2 + (x + 1)^2 + (x+2)^2 = (x + 3)^2 + (x +4)^2

Expand and solve quadratically.

3. This is kinda equivalent probabilistically to throwing two coins for one heads or more, or 1 - (Probability of getting a tails on BOTH of the other two coins)
 
answer for number 3

If three quarters are tossed and if at least one lands tails up, what is the probability that at least one lands heads up?

If we toss 3 quarters, there are 2^3 or 8 different ways they can land. HHH is the only combination with no tails, so there are 7 combinations with at least one tail. Of those 7 combinations with at least one tail, only TTT has no heads, so the rest have at least one head. The answer is therefore 6/7 or 85.7142857%.
 
Clarification requested

What does this expression mean -----> 2^3

Specifically, what does the carrot character mean, in this relationship, that is?

Thanks.
 
It means 2 to the power of 3. The carrot is supposed to indicate that the 3 should be superscript.
 
Clarifcation

So shall we say 2^3 = 8?

That is, 2 x 2 x 2?
 
You got it :smile:
 
probability

If three quarters are tossed and if at least one lands tails up, what is the probability that at least one lands heads up?

If we toss 3 quarters, there are 2^3 or 8 different ways they can land. HHH is the only combination with no tails, so there are 7 combinations with at least one tail. Of those 7 combinations with at least one tail, only TTT has no heads, so the rest have at least one head. The answer is therefore 6/7 or 85.7142857%.

How wrong is that! Even though HHH is not the event we are looking for, it can still happen and thus must be included in our calculation. The question maybe easier to understand if reformulated. What is the probability in 3 coin tosses of obtaining at least one H and one T. The only unwanted outcomes are HHH and TTT and thus the answer is 6/8. Or looking at it the other way if one coin definitely lands tails, then we have t consider 2 coin tosses, and find the probability of throwing at least one head. The only excluded outcome is TT and so the answer is 3/4, same answer. You cannot omit events as was done.:wink:
 
You cannot omit events as was done

Try this new problem:

"If I flip three coins and at least one of them is tails. What are the odds that they're all heads?"

Clearly zero, but your argument would say 1 in 8.


Semantics is a big killer in probability problems. The problem is asking for a conditional probability, but you were deriving a joint probability.

Symbolically:

A := "At least one is tails"
B := "At least one is heads"

the problem is asking for the probability of B given A is true:

P(B|A) = 6/7

You found the probability that both B and A are true:

P(B & A) = 3/4
 
Last edited:
  • #10
quite right, I stand corrected.
 

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