Question

3. Flip a coin 3 times, then record the number of heads in the table below. Repeat this procedure 25 times. Roll Value 1 2 3 4 5 6 7 8 9 10 11 12 13 Roll Value 14 15 16 17 18 19 20 21 22 23 24 25 4. Determine the experimental probability of obtaining: I a. exactly 2 heads b. all tails c. more tails than heads

          3. Flip a coin 3 times, then record the number of heads in the table below. Repeat this procedure 25 times.
Roll
Value
1
2
3
4
5
6
7
8
9
10
11
12
13
Roll
Value
14
15
16
17
18
19
20
21
22
23
24
25
4. Determine the experimental probability of obtaining:
I
a. exactly 2 heads
b. all tails
c. more tails than heads
        
Show more…
3. Flip a coin 3 times, then record the number of heads in the table below. Repeat this procedure 25 times.
Roll
Value
1
2
3
4
5
6
7
8
9
10
11
12
13
Roll
Value
14
15
16
17
18
19
20
21
22
23
24
25
4. Determine the experimental probability of obtaining:
I
a. exactly 2 heads
b. all tails
c. more tails than heads

Added by Amy W.

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Flip a coin 3 times, then record the number of heads in the table below. Repeat this procedure 25 times. Determine the experimental probability of obtaining: a. exactly 2 heads b. all tails c. more tails than heads 3. Flip a coin 3 times, then record the number of heads in the table below. Repeat this procedure 25 times. 2 3 5 6 7 8 9 10 11 12 13 Roll 1 4 Value 18 19 20 21 22 23 24 25 Roll Value 14 15 16 17 4. Determine the experimental probability of obtaining: a. exactly 2 heads b. all tails c. more tails than heads
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Transcript

-
00:01 In this problem, we are going to use various rules of probability in order to find the probabilities of certain events.
00:09 So in this question, we are given the percentage of different colors of candies.
00:17 Now, in the first problem, we need to find the probability of getting a green candy or a blue candy.
00:34 Now, first of all, we need to determine whether getting a green candy and getting a blue candy, whether these outcomes are mutually exclusive or not.
00:44 Now, the answer to this is yes, they are mutually exclusive because when one candy is chosen at random, it cannot be both green and blue.
00:54 So that means either it's going to be a green candy or it's going to be a blue candy or it's going to be neither of them.
01:00 But it cannot be green and blue at the same time.
01:03 So, both of these outcomes cannot happen simultaneously, hence the outcomes are mutually exclusive.
01:11 Because of this, using the addition rule, we can say that the probability of getting a green candy or a blue candy is the probability of getting a green candy plus the probability of getting a blue candy.
01:29 Now, according to the table that has been given, the green candies form 10 % of the candy.
01:36 Is in the back so the probability is 10 % and similarly blue candies also form 10 % of all the candies so the probability is once again 10%.
01:46 So 10 % plus 10 % is 20 % or you can write that as 220 by 100 which is equals to 0 .2.
01:57 So the probability is 20 % or 0 .2.
02:03 Similarly in the second question we need to find the probability of getting a yellow candy or a red candy.
02:15 Similar to the previous part, we need to first of all determine whether the outcomes are mutually exclusive.
02:21 In this case as well, they are mutually exclusive because on choosing one candy at random, it cannot be both yellow and red.
02:28 So both of these outcomes cannot happen together.
02:31 The outcomes are mutually exclusive...
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