Just a book problem I need help on:
A letter is picked at random from the alphabet. Find the probability that the letter is contained in the word "house" or in the word "phone".
I know this problem has something to do with mutually inclusive?? Would be great if someone could explain this to me
Thanks!
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$\begingroup$Hint: The set of letters in house is $\{h, o, u, s, e\}$; the set of letters in phone is $\{p, h, o, n, e\}$. You want to count the number of letters in the union of those two sets. In general, the number of elements in the union of sets $A$ and $B$ is $|A \cup B| = |A| + |B| - |A \cap B|$.
$\endgroup$ $\begingroup$I may have found the answer now?:
So the probability of 2 mutually inclusive events = P(A or B)= P(A) + P(B) - P(A and B). Not sure if that's the way you guys do it, but that's the way my book teaches me.
The probability of the letter being in the word "house" is 5/26
The probability of the letter being in the word "phone" is 5/26 also.
The probability of the letter being in both "phone" and "house" is 3/26, since there are 3 letters, being o, u, and h.
So, (5/26) + (5/26) - (3/26) is equal to 7/26. That's my answer. Please correct me if I'm wrong!
Thanks for the help guys.
$\endgroup$ 2 $\begingroup$Two events $A$ and $B$ are mutually exclusive iff they cannot both be true. To explain your question in terms of this concept, we could let the events
- $A =$ letter chosen is part of "house"
- $B =$ letter chosen is part of "phone"
Then the events $A$ and $B$ are not mutually exclusive, because both events could simultaneously happen (ie. if the letter chosen is an o, h or e).
So, to calculate the probability of $A$ OR $B$ happening, use: $P(A \text{ or } B)=P(A)+P(B)-P(A \text{ and } B)$, where $P(A \text{ and } B)$ is the probability of both $A$ and $B$ happening. If you don't already know this formula, you should look it up and understand why it is true.
$P(A)=5/26$ since there are $5$ distinct letters in "house" and $26$ possible letters to chose from. Similarly, $P(B)=5/26$.
$A$ and $B$ both happen iff the letter chosen is in both words - ie. the letter is o, h or e. So, $P(A \text{ and } B)=3/26$.
So $P(A \text{ or } B)=5/26 + 5/26 - 3/26 = 7/26$.
That answer can also be arrived at by considering the number of possible ways to get a letter that is in one or both of the words.
$\endgroup$ 2 $\begingroup$If you're not too comfortable yet with these probability problems, each time you work one you might want to back up and think about what the exact model for the problem is.
In this case:
Experiment: Pick one letter from the alphabet at random.
Sample space: There are 26 possible outcomes. $S = \{a, b, c, \ldots, z\}$.
Probability distribution: The problem is implicitly assuming all choices are equally likely here, so we have $P(\{\text{letter}\}) = 1/26$ for every letter.
Event: The letter you've chosen is contained in the word "house" or "phone." So $E = \{x \in S \mid x \text{ is one of } h, o, u, s, e, p, n \}$. That is, $E$ is just the set $\{h, o, u, s, e, p, n\}$. (You can break this down into a union of two separate events, as other answers are suggesting, also.)
The problem wants to know $P(E)$, which equals $P(\{h\}) + P(\{o\}) + \cdots + P(\{n\})$.
$\endgroup$ 1 $\begingroup$Let
$A$ = event that letter is in "house"
$\mathrm{P}(A) = \frac{5}{26}$
$B$ = event that letter is in "phone"
$\mathrm{P}(B) = \frac{5}{26}$
$A\cap B^c$ = the letter is in house but not phone (which is "u" and "e")
$\mathrm{P}(A \cup B^c) = \frac{2}{26}$
$B\cap A^c$ = the letter is in phone but not house (which is "p" and "n")
$\mathrm{P}(B\cap A^c) = \frac{2}{26}$
$A\cap B$ = the letter is in house and phone (which is "h" "o" "e")
$\mathrm{P}(A\cap B) = \frac{3}{26}$
$A \cup B$ = the letter is in $A$ or $B$
$\mathrm{P} \left( A \cup B \right) = \mathrm{P} \left( A \right)+\mathrm{P} \left( B\right) - \mathrm{P} \left( A \cap B \right) = \mathrm{P} \left( A \cap B^c \right) + \mathrm{P} \left( B \cap A^c \right) + \mathrm{P} \left( A \cap B \right) $
or simply count how many unique letters appear in both "house" and "phone" and devide by 26...
$\endgroup$ $\begingroup$Find the probability that the letter is contained in the word "house" or in the word "phone".
In short, find the probability that the letter is one from "e", "h", "n", "o", "p", "s" or "u".
I know this problem has something to do with mutually inclusive??
Indeed it has something to do with that. They are not mutually exclusive events; so the probability of the union is the sum of the probabilities of the events minus the probability of the intersection.
$$\Bbb P(X\in \text{"house"$\cup$"phone"}) \\=\\ \Bbb P(X\in \text{"house"})+\Bbb P(X\in \text{"phone"})-\Bbb P(X\in \text{"house"$\cap$"phone"})$$
There are five letters in the word "house", five letters in the word "phone", and three letters that are common to both words.
Put it together.
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