Asya and Vasya are cutting rectangles out of graph paper. Vasya is lazy; he rolls a die once and cuts out a square with a side equal to the number of points that come up. Asya rolls the die twice and cuts out a rectangle with length and width equal to the numbers that come up. Whose expected area of the rectangle is greater?
Let $A$ and $B$ be independent random variables taking values $1, 2, \ldots, 6$ with equal probabilities. Then the expected value of the area of A's rectangle is $\mathrm{E} A B = \mathrm{E} A \cdot \mathrm{E} B = (\mathrm{E} A)^2$ (since $A$ and $B$ are independent). The expected value of the area of Vasya's rectangle is $\mathrm{E} A^2$.

Since $0 < (\mathrm{E} A)^2 < \mathrm{E} A^2$.

Thus, on average, the area of Vasya's square is greater than the area of A's rectangle.
[ [discrete distribution ]

![](https://cdn.mathpix.com/cropped/2024_05_06_bf0205b4cbbe4b58a1adg-18.jpg?height=125&width=989&top_left_y=266&top_left_x=0) ..... Difficulty: $3+$

A coin is tossed 10 times. Find the probability that no two heads appear consecutively.

## Solution

The total number of outcomes when a coin is tossed 10 times is $2^{10}$. We need to find the number of combinations where no two heads appear consecutively.

First method. If there are no heads at all, then the sequence consists of ten tails, and there is only one such sequence. If there is one head, then there are 10 such combinations (the head can be in any of the 10 positions). If there are two heads, then the number of combinations is $C_{9}^{2}$ (we count the number of ways to place 2 heads among 8 tails or at the ends). And so on. If there are $k$ heads, then the number of combinations is $C_{11-k}^{k}$ (the number of ways to place the heads in $11-k$ positions among the tails and at the ends).

Thus, the total number of combinations is $1 + C_{10}^{1} + C_{9}^{2} + C_{8}^{3} + C_{7}^{4} + C_{6}^{5} = 144$.

Second method. Let the coin be tossed $n$ times, and let $f(n)$ be the number of ways to toss the coin without two heads appearing consecutively. The number of valid combinations where the last position is a tail is $f(n-1)$. The number of valid combinations where the last position is a head is $f(n-2)$, since a tail must be in the second-to-last position before the head. Thus, $f(n) = f(n-1) + f(n-2)$. Since $f(1) = 2$ and $f(2) = 3$, we can sequentially calculate $f(3) = 5$, $f(4) = 8$, ..., $f(10) = 144$.

Therefore, the desired probability is $144 : 2^{10} = 9 / 64$.

## Answer

$9 / 64$.