Does anyone know how to do this/have the answer? PLEASE
1 answer:
15: You solve problems like this by finding the probability of each case, and then multiplying them all. For each of the last 4 question, she has probability 1/2 of guessing right. So, she guessed 4 consecutive questions with probability
16: Like before: you have pick a king with probability 4/52 = 1/13 (there are four kings - one for each suit, out of 52 cards in a standard deck), and you pick "I" from "INCREDIBLE" with probability 2/10 = 1/5 (there are two "I"s out of 10 letters). So, the probability of picking a king and then an "I" is
17-20: The important bit of information here is that you replace the first ball. So, the first and second pick follow the exact same probability distribution, because they basically are two repetitions of the same experiment. So, for example, in ex. 17, the first ball is even with probability 15/30 = 1/2 (there are 15 even balls out of 30). Then, you have again probability 15/30 = 1/2 to pick an odd ball (there are also 15 odd balls out of 30). So, the probability of picking an even ball, replace it, and pick an odd ball is
Exercises 18 to 20 follow the same scheme: find out the probability of the two events and multiply them.
21-26: Not we DON'T replace the balls, so the second pick will suffer the effects of the first one. Let's dive into ex. 21 for example. For the first pick, we want a 2-digits number. There are 21 of such balls (all balls except balls 1 to 9), so we pick a 2-digits ball with probability 21/30 = 7/10. For the second pick, we want the balls number 4. But we have to assume that we already picked the first ball, and we picked a 2-digits ball. So, for the second pick, we're choosing from a bag with 29 balls, and there is only one ball labeled 4. So, we pick the ball number 4 with probability 1/29. We deduce that the two events happen one after the other with probability
Exercises 22 to 26 are similar: you find out the probability of the first event, and then you consider the new environment (i.e. you keep track of the first pick) when it comes to the probability distribution for the second pick).
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Part 1
The graph has even symmetry. You can see that because it is symmetric with respect to the y-axis.
Functions that have even symmetry have the following property:
Part 2
To answer this we can simply check if the property we mentioned earlier holds for this function.
We can see that sine does not have even symmetry.
In fact, sine function has the following property:
This is called odd symetry.
Part 3
Take a look at the function that you attached in the picture. We know that function has even symmetry.
Reflection over x-axis and <span>180° rotation around the origin would give us -f(x). We would not end up with the same function, so these two are out.
</span><span>90° rotation around the origin would mean we swapped x <span>and y</span> so that one is out too. R</span><span>eflection over the line y=x is a property of functions that have an odd symmetry.
We are left with reflection around y-axis and, as mentioned before, this is the property of evenly symmetric functions.</span>
The graph has even symmetry. You can see that because it is symmetric with respect to the y-axis.
Functions that have even symmetry have the following property:
Part 2
To answer this we can simply check if the property we mentioned earlier holds for this function.
We can see that sine does not have even symmetry.
In fact, sine function has the following property:
This is called odd symetry.
Part 3
Take a look at the function that you attached in the picture. We know that function has even symmetry.
Reflection over x-axis and <span>180° rotation around the origin would give us -f(x). We would not end up with the same function, so these two are out.
</span><span>90° rotation around the origin would mean we swapped x <span>and y</span> so that one is out too. R</span><span>eflection over the line y=x is a property of functions that have an odd symmetry.
We are left with reflection around y-axis and, as mentioned before, this is the property of evenly symmetric functions.</span>
Answer and explanation:
To find : List the sample space and tell whether you think the events are equally likely ?
Solution :
a) Toss 2 coins; record the order of heads and tails.
Let H is getting head and t is getting tail.
When two coins are tossed the sample space is {HH,HT,TH,TT}.
Total number of outcome = 4
As the outcome HT is different from TH. Each outcome is unique.
Events are equally likely since their probabilities are same.
b) A family has 3 children; record the number of boys.
Let B denote boy and G denote girl.
If there are 3 children then the sample space is
{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}
The possible number of boys are 0,1,2 and 3.
Number of boys Favorable outcome Probability
0 GGG
1 GGB,GBG,BGG
2 GBB,BGB,BBG
3 BBB
Since the probabilities are not equal the events are not equally likely.
c) Flip a coin until you get a head or 3 consecutive tails; record each flip.
Getting a head in a trial is dependent on the previous toss.
Similarly getting 3 consecutive tails also dependent on previous toss.
Hence, the probabilities cannot be equal and events cannot be equally likely.
d) Roll two dice; record the larger number
The sample space of rolling two dice is
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Now we form a table that the number of time each number occurs as maximum number then we find probability,
Highest number Number of times Probability
1 1
2 3
3 5
4 7
5 9
6 11
Since the probabilities are not the same the events are not equally likely.