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Emma and Josh have just gotten engaged. What is the probability that they have different blood types? Assume that blood types for both men and women are distributed in the general population according to the following proportions: $$ \begin{array}{cc} \hline \text { Blood Type } & \text { Proportion } \\ \hline A & 40 \% \\ B & 10 \% \\ A B & 5 \% \\ \mathrm{O} & 45 \% \\ \hline \end{array} $$

Short Answer

Expert verified
The probability that Emma and Josh have different blood types can be calculated using the complement rule. First calculate the probability of them having the same blood type and subtract that from 1 to get the probability that they have different blood types.

Step by step solution

01

Representation of probabilities

First we need to convert the given percentages into probabilities. The probability of blood type A is represented as 0.40, blood type B as 0.10, blood type AB as 0.05, and blood type O as 0.45.
02

Calculating the probability of the same blood type

Then we calculate the probability that Emma and Josh have the same blood type. The probability of them having the same blood type is the sum of the squares of the probabilities of each blood type. So that would be \((0.40)^2 + (0.10)^2 + (0.05)^2 + (0.45)^2\).
03

Calculating the probability of different blood types

The probability of Emma and Josh having different blood types is the complement of the probability of them having the same blood type. The total probability is 1 so: \[Prob(\text{different blood types}) = 1 - Prob(\text{same blood type})\] Calculating gives us the probability of them having different blood types.

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Key Concepts

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

Blood Type Distribution
In probability theory, the distribution of blood types is a fascinating example that demonstrates how certain traits are found in a population. Blood type distribution refers to how common each blood type is in a given group. In Emma and Josh's case, the distribution is as follows: blood type A makes up 40% of the population, B constitutes 10%, AB is 5%, and O is 45%. To find probabilities, these percentages are converted into decimal form: 0.40, 0.10, 0.05, and 0.45 respectively.

Understanding this distribution helps determine how likely it is for any two people to share or differ in blood type. These proportions indicate that blood type O is the most common while AB is the rarest. Such data can be helpful not just in calculating probability but also in medical contexts such as transfusions and disease studies.
Complementary Probability
Complementary probability involves finding the likelihood of an event by understanding the likelihood of its opposite occurring. In this exercise, we are determining the probability that Emma and Josh have different blood types by looking at the reverse - what if they had the same blood type?
  • First, calculate the probability they have the same blood type by squaring each probability and adding them together: \((0.40)^2 + (0.10)^2 + (0.05)^2 + (0.45)^2\).
  • This gives you the probability of Emma and Josh having the same blood type.
  • This is a classic example of complementary probability, where the probability of them having different blood types is simply one minus the probability of them having the same blood type.
With the principle of complementary probability, calculating the chance of differing blood types becomes straightforward, enhancing our understanding of how probabilities can be derived indirectly.
Probability Calculation
Calculating probability involves breaking down the event and determining its likelihood numerically. For Emma and Josh, once we understand the probabilities of blood types, we perform the necessary computations.
  • First, the probability they have the same blood type is determined by the sum \((0.40)^2 + (0.10)^2 + (0.05)^2 + (0.45)^2 = 0.42\).
  • The probability of them having different blood types is calculated as \(1 - 0.42 = 0.58\).
These calculations show that there is a 58% chance that Emma and Josh will have different blood types. This example emphasizes the simplicity and power of probability theory in analyzing real-world scenarios, showing that seemingly complex situations can be understood and quantified with clear numerical expressions.

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Most popular questions from this chapter

Building permits were issued last year to three contractors starting up a new subdivision: Tara Construction built two houses; Westview, three houses; and Hearthstone, six houses. Tara's houses have a \(60 \%\) probability of developing leaky basements; homes built by Westview and Hearthstone have that same problem \(50 \%\) of the time and \(40 \%\) of the time, respectively. Yesterday, the Better Business Bureau received a complaint from one of the new homeowners that his basement is leaking. Who is most likely to have been the contractor?

A man has \(n\) keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourceful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining \(n-1\) keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is \(1 / n\). Show that the probability the door opens with the third key he tries is also \(1 / n\). (Hint: What has to happen before he even gets to the third key?)

How many probability equations need to be verified to establish the mutual independence of four events?

A total of twelve hundred graduates of State Tech have gotten into medical school in the past several years. Of that number, one thousand earned scores of twenty-seven or higher on the MCAT and four hundred had GPAs that were \(3.5\) or higher. Moreover, three hundred had MCATs that were twenty-seven or higher and GPAs that were \(3.5\) or higher. What proportion of those twelve hundred graduates got into medical school with an MCAT lower than twenty-seven and a GPA below \(3.5\) ?

If two fair dice are tossed, what is the smallest number of throws, \(n\), for which the probability of getting at least one double 6 exceeds \(0.5 ?\) (Note: This was one of the first problems that de Méré communicated to Pascal in \(1654 .)\)

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