Your "naive first thought" is the clever (standard) solution. Show
To make it rigorous, let $\mathscr E_0$ be the event "A and B are tied after each has tossed 10 times;" let $\mathscr E_A$ and $\mathscr E_B$ be the events "A has more heads than B after 10 tosses each" and "B has more heads than A after 10 tosses each," respectively. Let $\mathscr F$ designate the event "A has more heads than B after all tosses are made." Notice:
By the law of total probability, $$\begin{aligned} \Pr(\mathscr F) &= \Pr(\mathscr F\mid \mathscr E_0)\Pr(\mathscr E_0) + \Pr(\mathscr F\mid \mathscr E_A)\Pr(\mathscr E_A) + \Pr(\mathscr F\mid \mathscr E_B)\Pr(\mathscr E_B)\\ & = \Pr(\mathscr E_0)\left(\frac{1}{2}\right) + \Pr(\mathscr E_A)(1) + \Pr(\mathscr E_B)(0)\\ &= \frac{1}{2}\left(\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_A)\right)\\ &= \frac{1}{2}\left(\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_B)\right)\\ & = \frac{1}{2}\left(1\right) = \frac{1}{2}. \end{aligned} $$ As an alternative approach, you wish to evaluate the double sum $$\sum_{a \gt b} \binom{11}{a}\binom{10}{b} = \sum_{a \gt b} \binom{11}{11-a}\binom{10}{10-b} = \sum_{a^\prime \le b^\prime} \binom{11}{a^\prime}\binom{10}{b^\prime}.$$ (In case the algebra isn't obvious, the first equality exploits the Binomial coefficient symmetry and the second is the change of variable $a^\prime = 11-a,$ $b^\prime = 10-b.$ We can be a vague about the endpoints of the summations because whenever $a$ or $a^\prime$ is not in the range from $0$ through $11$ or $b$ or $b^\prime$ is not in the range from $0$ through $10$ the Binomial coefficients are zero.) Because the indexes in the two sums on the left and right sides (1) never overlap and (2) cover all the possibilities (since either $a\gt b$ or $a\le b$ but never both), together they give the total probability, which is $1.$ Consequently, since those sums are equal, each is $1/2,$ QED.  This figure shows the rotational symmetry of the distribution under the mapping $(a,b)\to(11-a,10-b).$ The blue circles are rotated around the yellow dot into red triangles of exactly the same probability. The desired sum is the total of the blue circles, which therefore must be $1/2.$ A fair coin is flipped 11 times. Find the probability that more than 5 of the flips turn up tails. Expert Solution Students who’ve seen this question also like: Related Probability Q&AFind answers to questions asked by students like you. Q: At a car manufacturing plant, 20% of cars have some sort of defect when they come off the line. 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Knowledge Booster Learn more about Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below. Recommended textbooks for you College Algebra ISBN:9781337282291 Author:Ron Larson Publisher:Cengage Learning College Algebra ISBN:9781337282291 Author:Ron Larson Publisher:Cengage Learning What is the probability of flipping a coin and getting tails?Suppose you have a fair coin: this means it has a 50% chance of landing heads up and a 50% chance of landing tails up.
Which is more likely 9 heads in 10 tosses of a fair coin or 18 heads in 20 tosses?The probability of success (p) that is getting a head is 0.5 as it is a fair coin. Therefore, the probability of 18 heads in 20 tosses of a fair coin is 0.0002. Therefore, it can be concluded that 9 heads in 10 tosses of a fair coin is more likely than 18 heads in 20 tosses of a fair coin.
What is the probability if I flip a fair coin with heads and tails ten times in a row that I get at least 88 heads?The answer is the probability, that out of ten tosses of the coin, at least 8 show heads, is 0.0547 .
What is the probability of getting exactly 5 heads in 10 coin flips?So, the probability of getting exactly 5 heads when 10 coins are tossed is 63256. Hence answer is 63256. Note: There is a possibility of making a mistake while finding the number of ways of arranging 5 heads and 5 tails.
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A fair coin is flipped 11 times find the probability that more than 7 of the flips turn up tails
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