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 $1/2.$ We build a mathematical model of the experiment. Write H for head and T for tail. Record the results of the tosses as a string of length $10$, made up of the letters H and/or T. So for example the string HHHTTHHTHT means that we got a head, then a head, then a head, then a tail, and so on. Nội dung chính Show
There are $2^{10}$ such strings of length $10$. This is because we have $2$ choices for the first letter, and for every such choice we have $2$ choices for the second letter, and for every choice of the first two letters, we have $2$ choices for the third letter, and so on. Because we assume that the coin is fair, and that the result we get on say the first $6$ tosses does not affect the probability of getting a head on the $7$-th toss, each of these $2^{10}$ ($1024$) strings is equally likely. Since the probabilities must add up to $1$, each string has probability $\frac{1}{2^{10}}$. So for example the outcome HHHHHHHHHH is just as likely as the outcome HTTHHTHTHT. This may have an intuitively implausible feel, but it fits in very well with experiments. Now let us assume that we will be happy only if we get exactly $3$ heads. To find the probability we will be happy, we count the number of strings that will make us happy. Suppose there are $k$ such strings. Then the probability we will be happy is $\frac{k}{2^{10}}$. Now we need to find $k$. So we need to count the number of strings that have exactly $3$ H's. To do this, we find the number of ways to choose where the H's will occur. So we must choose $3$ places (from the $10$ available) for the H's to be. We can choose $3$ objects from $10$ in $\binom{10}{3}$ ways. This number is called also by various other names, such as $C_3^{10}$, or ${}_{10}C_3$, or $C(10,3)$, and there are other names too. It is called a binomial coefficient, because it is the coefficient of $x^3$ when the expression $(1+x)^{10}$ is expanded. There is a useful formula for the binomial coefficients. In general $$\binom{n}{r}=\frac{n!}{r!(n-r)!}.$$ In particular, $\binom{10}{3}=\frac{10!}{3!7!}$. This turns out to be $120$. So the probability of exactly $3$ heads in $10$ tosses is $\frac{120}{1024}$. Remark: The idea can be substantially generalized. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ This probability model is called the Binomial distribution. It is of great practical importance, since it underlies all simple yes/no polling. Video TranscriptFor this exercise, we are told that coin is flipped 14 times and we know that it's a fair coin. So the probability of turning up tails is one half. We are asked to find the probability that more than six of the flips turn up tails. So let's define the random variable X as the number of tails Out of 14 flips. Now each flip can be viewed as a Bernoulli trial because each flip is independent from the other flips. So the outcome for one has no bearing on the others. The probability of success is always half. Now the number of successes in a series of Bernoulli trials is a binomial random variable. So here X is a binomial random variable and the binomial has two parameters, probability of success and and number of trials and so we want the probability of more than six tails Out of 14 flips. That is the probability that X is greater than six. Now for a binomial random variable, the probability mass function is given by this formula and choose X. That's peter the exponent X. Um so 1 -2 to the exponent and mind sex And X can take on any integer value from zero up to N. Now we can express this probability probability as one the probability that X Is at most six And then that would be 1- this summation. This isn't too bad to calculate by hand, but there's eight terms, they're seven or eight terms. So it's much more convenient to use software. Let's use Excel for this. So in Excel we can start by entering equals to start a calculation First we had 1- and then we want the probability that X is at most six. So we start to type binomial and this is the function that we want bynum dot d i s T. So we enter six successes. Number of trials is 14, Probability of success is .5. And then for a cumulative we select true because we want a cumulative probability We're looking for the probability that X is less than or equal to six and hit enter And we get .0.6047. So that is the answer. And that looks like it's the 4th multiple choice option in your question. Okay. I guess that's it. I hope this helped. And good luck in your studies.
What is the probability of flipping a coin 3 times and getting tails?Answer: The probability of flipping a coin three times and getting 3 tails is 1/8. What is the probability of getting no more than 3 heads when you flip a fair coin 5 times?Considering a fair coin, after 5 flips, there are 25 = 32 different arrangements of heads and tails. Therefore, the probability of exactly 3 heads is 5/16. What is the probability of getting 3 heads or less in 10 tosses of a fair coin?So the probability of getting exactly three heads-- well, you get exactly three heads in 10 of the 32 equally likely possibilities. So you have a 5/16 chance of that happening. What is the probability that when you flip a coin 3 times you will get 3 heads?Correct answer: If you flip a coin, the chances of you getting heads is 1/2. This is true every time you flip the coin so if you flip it 3 times, the chances of you getting heads every time is 1/2 * 1/2 * 1/2, or 1/8. 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.
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 obtaining six tails in a row when flipping a coin interpret this probability?And then we want to flip the coin six times and we want to get six tails in rule, which means that we have a tail for each trial. Then the probability is 1/2 to the power of six And then this is . 0156.
What is the probability of flipping heads 20 times in a row?The chance of n heads in a row occurring is 1/2n, so the inverse probability is (2n-1)/2n. If we multiply that probability once for all 999,981 possible occurences of a streak of 20 heads, it seemed to me that I would be in business. Doing this is a simple enough calculation, and the result was the 60% figure.
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A fair coin is flipped 13 times find the probability that more than 7 of the flips turn up tails
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