Question

An element is randomly chosen from among the first $15$ rows of pascal's triangle. What is the probability that the value of the element chosen is $1$? Note: the 1 at the top is often labelled the "zeroth" row of pascal's triangle, by convention. So to count a total of 15 rows, use rows 0 through 14.

Answer 1

Answer:

Probability = 29/120

Step-by-step explanation:

In pascal's triangle, number of elements in each raw are same as the number of the raw.

i.e.

in 1st raw lablled as 'zeroth' raw - number of elements  = 1

in 2nd raw, - number of elements = 2

.

.

in 15th raw - number of elements = 15

Thus total number of elements in first 15 rows of Pascal's trianle = 1+ 2 + ... 15

∴ Total number of elements = n×(n+1)÷2

∴ Total number of elements in 15 rows = 15×16÷2

∴ Total Number of elements in 15 rows = 120  ....... (1)

In pascal's triangle, every row starts and ends with '1'

∴ except 'Zeroth' row, every row have two elements with value '1'

& there is one element in 'zeroth' row whose value is '1'

∴ total number of elements in 15 rows with value '1' = 1 + 2×(15-1)

∴ total number of elements in 15 rows with value '1' = 29 ..... (2)

∴ Probability that value of element chosen is '1' = (Number of elements with value '1') ÷ (total number of elements in Pascal's triangle in 15 rows)

∴ Probability that value of element chosen is '1' = 29/120

Answer 2

Answer:

29/120

Step-by-step explanation:

First we find the total number of elements in the first 15 rows. The first row of Pascal's Triangle has one element, the second row has two, and so on. The first 15 rows thus have $1+2+\cdots+15$ elements. Instead of manually adding the summands, we can find the sum by multiplying the average of the first and last term $\frac{1+15}{2}$ by the number of terms, $15$. The sum is $\frac{16}{2}\cdot15=8\cdot15=120$, so there are $120$ elements. Now we find the number of ones in the first $15$ rows. Each row except the first has two ones, and the first row only has one. So there are $14\cdot2+1=29$ ones. With $29$ ones among the $120$ possible elements we could choose, the probability is $\boxed{\frac{29}{120}}$.