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STALIN [3.7K]
3 years ago
7

There is a 20% chance that it will snow on Monday and a 40% chance on Tuesday. What is the probability that it will NOT snow on

both Monday and Tuesday?
Mathematics
2 answers:
andrezito [222]3 years ago
8 0
. Solve: 5x < 9x-1 / 2 

<span>-4x < - 1/2 , now divide by -4 and switch signs </span>

<span>x > 1/8 </span>


<span>2. There is a 20% chance that it will snow on Monday and a 40% chance on Tuesday. What is the probability that it will NOT snow on both Monday and Tuesday? </span>

<span>P( NOT snow on both Monday and Tuesday) = (1-.2)(1-.4) = 12/25 </span>

<span>c. 12/25 </span>


<span>3. The total cost of renting a sailboat is $120 plus $60 a day. How many days was a sailboat rented if the total cost was $360? Solve. </span>

<span>120 + 60d = 360 </span>

<span>60d = 360 - 120 = 240 </span>

<span>d = 4 </span>



<span>4. Two cards are chosen at random without replacement from a deck of 52 cards containing 4 kings. Which expression could be used to find the probability of choosing two kings? </span>

<span>b. 4/52 x 3/51 </span>

<span>There are 4 x 3 ways to pick 2 kings </span>

<span>There are 52 x 51 ways to pick any two cards</span>

lapo4ka [179]3 years ago
7 0
P(Event) = 4/5 * 3/5 = 12/25

----------------------

Y=Yes=Will Snow

N=No=Won't Snow

There is a 20% chance (1/5) that it will snow on Monday. This means that there is an 80% (4/5) chance that it won't snow on Monday.

Y, N, N, N, N

There is a 40% chance (2/5) that it will snow on Tuesday. This means that there is a 60% chance (3/5) that it won't snow on Tuesday.

Y, Y, N, N, N

-----------------------

Create a table for combinations N, N or Y, N or N, Y or Y, Y:

*Where N, N place (X). Where Y, N or N, Y or Y, Y - place (O).

   : Y  N  N  N  N
.................................
Y : O  O  O  O O
Y : O  O  O  O O
N : O  X  X  X  X
N : O  X  X  X  X
N : O  X  X  X  X

The combination N, N represented with (X) appears 12 times out of 25 which is your answer (12/25).
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Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not  injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true  that f must be surjective? Is it true that g must be surjective? Justify your answers with either a  counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not  injective (since both 2, 3 7→ 4) but g ◦ f is injective.  Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2  . Then g is not  injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t  specify the domain and codomain though. Note that the codomain of f must equal the domain of

g for g ◦ f to make sense.

(2) Answer

Solution: There are two questions in this problem.

Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}.  Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that  g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective.  Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we  must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the  c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).

Then g(b) = g(f(a)) = c and we have found our desired b.  Remark: It is good to compare the answer to this problem to the answer to the two problems

on the previous page.  The part of this problem most groups had the most issue with was the second. Everyone should  be comfortable with carefully proving a function is surjective by the time we get to the midterm.

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