## Coin Toss

Consider 10 coins are tossed simultaneously. What is the probability of getting at least 9 tails? Assume either heads or tails for only possible outcomes.

##
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__**Hint**

**Hint**

There are exactly two possible outcomes (heads/tails) for every coin toss. Regardless if thrown simultaneously or not, there will be a one in two chance of the coin landing tails. Thus, the total number of possible outcomes is:

$$$\frac{1}{2}\cdot \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2^{10}}=\frac{1}{1024}$$$

##
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__**Hint 2**

**Hint 2**

Unlike a permutation where a particular sequence order is considered, a combination is needed to find the number of favorable events (i.e. the number of outcomes in which there are 9 or 10 tails) because all the coins are tossed simultaneously.

There are exactly two possible outcomes (heads/tails) for every coin toss. Regardless if thrown simultaneously or not, there will be a one in two chance of the coin landing tails. Thus, the total number of possible outcomes is:

$$$\frac{1}{2}\cdot \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2^{10}}=\frac{1}{1024}$$$

Unlike a permutation where a particular sequence order is considered, a combination is needed to find the number of favorable events (i.e. the number of outcomes in which there are 9 or 10 tails) because all the coins are tossed simultaneously:

$$$C_{(n,r)}=\frac{P_{(n,r)}}{r!}=\frac{n!}{[r!(n-r)!]}$$$

where
$$C_{n,r}$$
is the number of different combinations of
$$n$$
distinct objects taken
$$r$$
at a time, and
$$P$$
is the number of different permutations. Thus, the number of favorable outcomes for our problem statement is:

$$$C_{(10,9)}+C_{(10,10)}$$$

which is the number of ways to select 9 coins out of 10 plus the number of ways to select 10 coins out of 10.

$$$=\frac{10!}{[9!(10-9)!]}+\frac{10!}{[10!(10-10)!]}=\frac{10!}{9!\cdot 1!}+\frac{10!}{10!\cdot 0!}=\frac{10}{1}+\frac{1}{1}=11$$$

Finally, the probability of getting at least 9 tails:

$$$\frac{11}{1024}$$$

$$$\frac{11}{1024}$$$