Theoretical Probabilities of Coin Tosses (Part 2) hink of tossing a coin 10 times as a binomial probability experiment, where the probability of heads is p = 0.5. Compute the remaining theoretical probabilities of obtaining a certain number of heads in 10 tosses in the table elow* Enter the results from your empirical trials in the last two columns. (Ex: the empirical probability of obtaining exactly 4 heads in 10 tosses is the proportion of your twelve trials that resulted in obtaining 4 heads.) No. of heads 0 1 2 3 4 No. of combinations 10! 0! (10-0)! 10! 1! (10-1)! 10! 2! (10-2)! G 10! 3! (10-3)! = 1 =10 = 45 = 120 10! 3! (10-3)! = 120 Probability of an individual outcome 1 2 1 2 1 2 HIN 1 10 ¹× (-)* ×(₁-)" = 1 x 1- Theoretical Probability 45 1 10 × () ×(1--) = 120 1- = 0.001 8 = 0.01 = 0.044 = 0.117 Frequency of trials with this No. of heads Empirical Probability of this No. of heads
Theoretical Probabilities of Coin Tosses (Part 2) hink of tossing a coin 10 times as a binomial probability experiment, where the probability of heads is p = 0.5. Compute the remaining theoretical probabilities of obtaining a certain number of heads in 10 tosses in the table elow* Enter the results from your empirical trials in the last two columns. (Ex: the empirical probability of obtaining exactly 4 heads in 10 tosses is the proportion of your twelve trials that resulted in obtaining 4 heads.) No. of heads 0 1 2 3 4 No. of combinations 10! 0! (10-0)! 10! 1! (10-1)! 10! 2! (10-2)! G 10! 3! (10-3)! = 1 =10 = 45 = 120 10! 3! (10-3)! = 120 Probability of an individual outcome 1 2 1 2 1 2 HIN 1 10 ¹× (-)* ×(₁-)" = 1 x 1- Theoretical Probability 45 1 10 × () ×(1--) = 120 1- = 0.001 8 = 0.01 = 0.044 = 0.117 Frequency of trials with this No. of heads Empirical Probability of this No. of heads
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Transcribed Image Text:Theoretical Probabilities of Coin Tosses (Part 2)
Think of tossing a coin 10 times as a binomial probability experiment, where the probability of heads is p = 0.5.
Compute the remaining theoretical probabilities of obtaining a certain number of heads in 10 tosses in the table
below*
Enter the results from your empirical trials in the last two columns. (Ex: the empirical probability of obtaining
exactly 4 heads in 10 tosses is the proportion of your twelve trials that resulted in obtaining 4 heads.)
+
No. of
heads
0
2
3
5
6
7
8
9
10
No. of
combinations
10!
0! (10-0)
10!
1! (10-1)!
<<= 10
10!
2: (10-2)!
= 1
10!
3! (10-3)!
= 45
<= 120
10!
3! (10-3)!
<= 120
Probability
of an
individual
outcome
13
1
2
HİN
2
12
IN
15
13
2
10 x
Theoretical Probability
× (²) ×(1-1) = 0.
120
10
8
= 0.001
2
45 ()*(1-3)* -0.044
7
0.01
= 0.117
Frequency
of trials
with this
No. of
heads
Empirical
Probability
of this No.
of heads
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