Solving Probability Problem: Flip a Coin n Times (X) and n+1 Times (Y)

• MHB
• jamcc09
In summary: Y) = \frac{1-0.5^n}{2}$.It is possible to use part (a) in part (c), but it is not necessary. The key is to recognize that$P(X<Y) + P(X>Y) + P(X=Y) = 1$. From this, we can solve for$P(X<Y)$using the fact that$P(X>Y) = P(Y<X) = P(n-Y<n-X) = P(n-X<n+1-Y)$. This leads to the same result as before,$P(X<Y) = \frac{1-P(X=Y)}{2}$.I hope this helps clarify your questions. Keep up the good work jamcc09 As you may notice from a previous post of mine, I am working through the Blitzstein & Hwang probability book. I have run into a problem, which I cannot quite solve. It goes like this: Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent$X \sim Bin(n, 1/2)$and$Y \sim Bin(n+1, 1/2)$. (a) Let$V = min(X,Y)$and$W = max(X,Y)$. Find$E(V) + E(W)$. If$X=Y$then$X=Y=V=W$. (b) Show that$P(X<Y) = P(n-X<n+1-Y)$. (c) Compute$P(X<Y)$. Hint: Use (b) and the fact that$X$and$Y$are integer-valued. In case anyone else is interested (a) is answered simply by seeing that$X+Y = V+W$and therefore$E(V+W)=E(X+Y)=E(X)+E(Y)=E(V)+E(W)$. This simply uses linearity of expectation. For (b), I believe that$P(X<Y) = P(n-X<n+1-Y)$by symmetry, since both$X$and$Y$are binomially-distributed with$p=0.5$. So, my first question is: is this correct? I am not really sure how to justify this mathematically, but seems to be intuitively true because$P(X=k) = P(X=n-k)$(although maybe this is wrong ;)). For (c), the question suggests using part (b). So, I took that to mean$1 = P(X<Y) + P(Y>X) + P(X=Y)$. So, I took that to mean $$1 = P(X<Y) + P(Y>X) + P(X=Y) = P(X<Y) + P(n-X<n+1-Y) + P(X=Y) = 2P(X<Y) + P(X=Y).$$Therefore,$P(X<Y) = \frac{1-P(X=Y)}{2}$. I think then$P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k)$. I feel like I am now using that$X$and$Y$are integer-valued, but I am thinking there is a further simplification of this formula. Can someone help me get on the right track here? Just an additional note, is it possible to use part (a) at all in (b) and (c)? I don't really see how these are connected. Thanks, - J Last edited: Dear J, Thank you for sharing your progress and questions with us. It seems like you are making good progress in understanding the material in the Blitzstein & Hwang probability book. I can offer some insights and guidance on your questions. (a) Your solution to this part is correct and well-reasoned. As you noted, this is a simple application of the linearity of expectation, which states that the expected value of a sum of random variables is equal to the sum of their expected values. (b) Your intuition is correct. The symmetry of the binomial distribution with$p=0.5$implies that$P(X=k) = P(X=n-k)$, and similarly for$Y$. This can be justified mathematically by the symmetry of the binomial probability mass function, which is$P(X=k) = P(X=n-k)$for all integer values of$k$. Therefore,$P(X<k) = P(X>n-k) = P(n-X<k)$, and similarly for$Y$. This leads to the desired result$P(X<Y) = P(n-X<n+1-Y)$. (c) Your approach is correct, but there is a simpler way to compute$P(X=Y)$using the binomial distribution. Since$X$and$Y$are independent, the joint probability of$X$and$Y$is given by the product of their individual probabilities. Therefore,$P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k) = \sum_{k=0}^{n} P(X=k)P(X=k) = \sum_{k=0}^{n} P(X=k)^2$. Since$X$is binomially distributed with$p=0.5$, we have$P(X=k) = \binom{n}{k}0.5^k0.5^{n-k} = \binom{n}{k}0.5^n$. Substituting this into the sum, we get$P(X=Y) = \sum_{k=0}^{n} \binom{n}{k}0.5^{2n} = 0.5^{2n}\sum_{k=0}^{n} \binom{n}{k} = 0.5^{2n}(2^n) = 0.5^n$. Therefore,$P(X

What is the probability of getting exactly X heads when flipping a coin n times?

The probability of getting exactly X heads when flipping a coin n times is given by the binomial distribution formula: P(X) = (n choose X) * (0.5)^X * (0.5)^(n-X), where n is the number of flips and X is the number of desired heads. This formula assumes that the coin is fair and each flip is independent of the previous ones.

How does the probability of getting X heads change as n increases?

As n increases, the probability of getting X heads approaches a normal distribution with a mean of n/2 and a standard deviation of sqrt(n/4). This means that as n gets larger, the probability of getting X heads gets closer to 50%. For example, if n = 10, the probability of getting exactly 5 heads is approximately 24.6%, whereas if n = 100, the probability of getting exactly 50 heads is approximately 7.96%.

What is the probability of getting more than X heads when flipping a coin n times?

The probability of getting more than X heads when flipping a coin n times can be calculated by adding the individual probabilities of getting X+1, X+2, ..., n heads. This can be simplified to 1 - P(X), where P(X) is the probability of getting exactly X heads. For example, if n = 5 and X = 2, the probability of getting more than 2 heads is 1 - 0.3125 = 0.6875.

What is the expected value of the number of heads when flipping a coin n times?

The expected value of the number of heads when flipping a coin n times is n/2. This means that on average, we can expect to get n/2 heads when flipping a coin n times. This is because the probability of getting a head is 0.5, and we are flipping the coin n times, so the expected value is n * 0.5 = n/2.

How does the probability of getting X heads change when flipping the coin n+1 times compared to n times?

The probability of getting X heads when flipping the coin n+1 times is the same as the probability of getting X heads when flipping the coin n times. This is because each flip is independent and the probability of getting a head or tail remains the same. The only difference is that there is an extra flip, but it does not affect the probability of getting X heads.

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