Question

Apply the binomial theorem to 210 = (1 + 1)10 . Answer each question in the work area provided below. 1. Write out the equation using combination notation. 2. Simplify each term. Do not combine the terms. 3. In two or more complete sentences, describe the relationship between the terms of the equation and the number of combinations when flipping a coin 10 times. 4. What is the theoretical probability of getting exactly 5 heads in a sequence of 10 flips? 5. In two or more complete sentences, compare the theoretical probability of getting exactly 5 heads with the fraction in your histogram. Interpret any differences.

Answer 1

Answer:

The following are the answer to this question:

Step-by-step explanation:

Binomial theorem Expression:

$\bold{(x+y)^n= {^n}C_0x^ny^0+{^n}C_1x^{n-1}y^1+{^n}C_2x^{n-2}y^2+...+{^n}C_{\gamma}x^{n-\gamma}y^\gamma}+...+ {^n}C_nx^0y^n$Solution:

1)

$\to 2^{10} = (1 + 1)^{10}$

         $= (1+1)^{10}= {^{10}}C_0\times 1^{10}\times 1^0+{^{10}}C_1\times 1^{9}\times 1^1+{^{10}}C_2\times 1^{8}\times 1^2+ \\ {^{10}}C_3 \times 1^{7}\times 1^3+{^{10}}C_4 \times 1^{6}\times 1^4+ {^{10}}C_5\times 1^{5}\times 1^5 \\+ {^{10}}C_6 \times 1^{4}\times 1^6+....+{^{10}}C_{10}\times 1^{0}\times 1^{10}$

          $= {^{10}}C_0. 1^{10}. 1^0+{^{10}}C_1.1^{9}. 1^1+{^{10}}C_2.1^{8}.1^2+\\{^{10}}C_3.1^{7}.1^3+{^{10}}C_4. 1^{6}. 1^4+ {^{10}}C_5.1^{5}. 1^5 + {^{10}}C_6 . 1^{4}. 1^6+ \\ {^{10}}C_7 . 1^{3}. 1^7 + {^{10}}C_8 . 1^{2}. 1^8+ {^{10}}C_9 . 1^{1}. 1^9+{^{10}}C_{10}. 1^{0}. 1^{10}$

2)

Simplify:

$(1+1)^{10}=$

${^{10}}C_0. 1^{10}. 1^0+{^{10}}C_1.1^{9}. 1^1+{^{10}}C_2.1^{8}.1^2+\\{^{10}}C_3.1^{7}.1^3+{^{10}}C_4. 1^{6}. 1^4+ {^{10}}C_5.1^{5}. 1^5 + {^{10}}C_6 . 1^{4}. 1^6+ \\ {^{10}}C_7 . 1^{3}. 1^7 + {^{10}}C_8 . 1^{2}. 1^8+ {^{10}}C_9 . 1^{1}. 1^9+{^{10}}C_{10}. 1^{0}. 1^{10}$

$=1+10+45+120+210+252+210+120+45+10+1\\\\=1024$

3)

The $r^{th}$ word is the number of variations where a coin is redirected 10 times.

4)

Get the frequency of exactly five heads is: ${^{10}}C_{5}$ = 252

        $=\frac{252}{1024}\\\\=\frac{126}{512}\\\\=\frac{63}{256}\\\\=0.24$

5)

To Get 5 heads will be:

$={^{10}}C_{5} \times (\frac{1}{2})^{10}\\\\=\frac{{^{10}}C_{5}}{2^{10}}\\\\=0.24$