Suppose George wins 38​% of all thumb wars. ​(a) What is the proba

pro4ph5e4q2

pro4ph5e4q2

Answered question

2021-11-24

Suppose George wins 38​% of all thumb wars.
​(a) What is the probability that George wins two thumb wars in a​ row?
​(b) What is the probability that George wins four thumb wars in a​ row?
​(c) When events are​ independent, their complements are independent as well. Use this result to determine the probability that George wins four thumb wars in a​ row, but does not win five in a row.

Answer & Explanation

Melinda Olson

Melinda Olson

Beginner2021-11-25Added 20 answers

Probability(win game) =p(W)=0.34
let W=w,L=lose
a) p(winning 2 wars simultaneously)=0.340.34=0.1156
b) p(winning 4 games simultaneously)=0.34×0.34×0.34×0.34=0.344=0.01336
c)events are​ independent, their complements are independent as well.
A W has 0.34 probability. A L has 0.66 probability
So p(winning 4 in a row and failure in the 5th)=0.344×0.66=0.00882
Don Sumner

Don Sumner

Skilled2023-06-19Added 184 answers

Step 1:
(a) To find the probability that George wins two thumb wars in a row, we multiply the probability of winning one thumb war by itself since the events are independent. Let P(G) denote the probability that George wins a thumb war. Therefore, the probability that George wins two thumb wars in a row is given by:
P(GG)=P(G)×P(G)=(0.38)×(0.38)=0.382=0.1444
Step 2:
(b) Similarly, to find the probability that George wins four thumb wars in a row, we multiply the probability of winning one thumb war by itself four times. Therefore, the probability that George wins four thumb wars in a row is given by:
P(GGGG)=P(G)×P(G)×P(G)×P(G)=(0.38)4=0.014084
Step 3:
(c) According to the given result, if events are independent, their complements are independent as well. Therefore, the probability that George wins four thumb wars in a row but does not win five in a row is given by:
P(GGGGG)=P(G)×P(G)×P(G)×P(G)×P(G)=(0.38)4×(10.38)=0.014084×0.62=0.00872288
Vasquez

Vasquez

Expert2023-06-19Added 669 answers

(a) The probability of George winning two thumb wars in a row can be calculated by multiplying the probability of him winning one thumb war with the probability of him winning the second thumb war, assuming independence.
Let's denote the probability of George winning a thumb war as P(win), and the probability of him losing as P(lose). Since the probability of winning is given as 38%, we have P(win)=0.38.
Using the independence assumption, the probability that George wins two thumb wars in a row is given by:
P(win two in a row)=P(win)×P(win)=0.38×0.38=0.1444
Therefore, the probability that George wins two thumb wars in a row is 0.1444 or 14.44%.
(b) Similarly, the probability that George wins four thumb wars in a row can be calculated using the same approach:
P(win four in a row)=P(win)×P(win)×P(win)×P(win)=0.38×0.38×0.38×0.38=0.0383
Therefore, the probability that George wins four thumb wars in a row is 0.0383 or 3.83%.
(c) According to the given information, if events are independent, their complements are also independent. This means that the probability of George not winning five thumb wars in a row is equal to the complement of him winning four thumb wars in a row.
Using this result, the probability that George wins four thumb wars in a row but does not win five in a row is:
P(win four, not five in a row)=P(win four in a row)×P(notwin)
The probability of not winning a thumb war is given by P(lose)=1P(win)=10.38=0.62.
Substituting the values, we have:
P(win four, not five in a row)=0.0383×0.62=0.0238
Therefore, the probability that George wins four thumb wars in a row but does not win five in a row is 0.0238 or 2.38%.
RizerMix

RizerMix

Expert2023-06-19Added 656 answers

Answer:
(a) 0.1444 or 14.44%
(b) 0.0383 or 3.83%
(c) 0.0112 or 1.12%
Explanation:
(a) The probability of George winning two thumb wars in a row can be calculated by multiplying the probability of winning a single thumb war by itself:
P(George wins two in a row)=P(George wins a single thumb war)×P(George wins a single thumb war)
Given that George wins 38% of all thumb wars, the probability of him winning a single thumb war is 38% or 0.38. Therefore:
P(George wins two in a row)=0.38×0.38=0.1444
So, the probability that George wins two thumb wars in a row is 0.1444 or 14.44%.
(b) Similarly, the probability of George winning four thumb wars in a row can be calculated by multiplying the probability of winning a single thumb war four times:
P(George wins four in a row)=P(George wins a single thumb war)×P(George wins a single thumb war)×P(George wins a single thumb war)×P(George wins a single thumb war)
P(George wins four in a row)=0.38×0.38×0.38×0.38=0.0383
So, the probability that George wins four thumb wars in a row is 0.0383 or 3.83%.
(c) According to the given statement, when events are independent, their complements are also independent. The complement of George winning four thumb wars in a row is George not winning four thumb wars in a row but winning the fifth one.
To find the probability of George winning four thumb wars in a row but not winning the fifth one, we can subtract the probability of winning five in a row from the probability of winning four in a row:
P(George wins four but not five in a row)=P(George wins four in a row)P(George wins five in a row)
Since each thumb war is assumed to be independent, the probability of George winning five thumb wars in a row is:
P(George wins five in a row)=P(George wins a single thumb war)5
P(George wins five in a row)=0.385=0.0271
Therefore, P(George wins four but not five in a row)=0.03830.0271=0.0112
So, the probability that George wins four thumb wars in a row but does not win five in a row is 0.0112 or 1.12%.

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