Most likely coin-tossing sequence. In Parade Magazine’s (November 26, 2000) column “Ask Marilyn,” the following question was posed: “I have just tossed a [balanced] coin 10 times, and I ask you to guess which of the following three sequences was the result. One (and only one) of the sequences is genuine.”

(1) H HHHHHHHHH

(2) H H T T H T T H HH

(3) T TTTTTTTTT

1. Demonstrate that prior to actually tossing the coins, thethree sequences are equally likely to occur.
2. Find the probability that the 10 coin tosses result in all heads or all tails.
3. Find the probability that the 10 coin tosses result in a mix of heads and tails.
4. Marilyn’s answer to the question posed was “Though the chances of the three specific sequences occurring randomly are equal . . . it’s reasonable for us to choose sequence (2) as the most likely genuine result.” If you know that only one of the three sequences actually occurred, explain why Marilyn’s answer is correct. [Hint: Compare the probabilities in parts b and c.]

Please choose one of the following three sequences as the outcome of the coin toss, I just completed. The only real sequence is one (and only one).

(1)$\text{HHHHHHHHHH}$

(2)$\text{HHTTHTTHHH}$

(3)$\text{TTTTTTTTTT}$

For an event A.

The require formula is $\mathrm{P}\left(\text{H}\cap \text{H}\cap \text{H}\cap \text{H}\cap \text{H}\cap \text{H}\cap \text{H}\cap \text{H}\cap \text{H}\cap \text{H}\right)$.

Thus

$\begin{array}{c}\mathrm{P}\left(\text{(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)}\right)=\frac{1}{1024}\\ =0.000977\end{array}$

For an event B

The require formula is$\mathrm{P}\left(\text{H}\cap \text{H}\cap \text{T}\cap \text{T}\cap \text{H}\cap \text{T}\cap \text{T}\cap \text{H}\cap \text{H}\cap \text{H}\right)$

Then

$\begin{array}{c}\mathrm{P}\left(\text{(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)(}0.50\text{)}\right)=\frac{1}{1024}\\ =0.000977\end{array}$

For event c.

The require formula is$\mathrm{P}\left(\text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\right)$

Thus

.$\mathrm{P}\left(\text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\cap \text{T}\right)$

Every coin toss offers 2 possible outcomes.Then the number of possible coin tossing

sequence for ten tosses is $\mathrm{P}\text{(}{2}^{10}\text{)}=1024$.

Since, I have number of alternative coins tossing sequence from ten tosses. Therefore,

$\frac{1}{1024}=0.000977$

The is probability is

$\begin{array}{c}\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{C}\text{)}=\mathrm{P}\text{(}\mathrm{A}\text{)}+\mathrm{P}\text{(}\mathrm{C}\text{)}\\ =0.000977+0.000977\\ =0.001954\end{array}$

The probability is

$\begin{array}{c}\mathrm{P}\text{(headsandtailsaremixedtogether)}=1-\mathrm{P}\text{(}\mathrm{A}\cap \mathrm{C}\text{)}\\ =1-0.001954\\ =0.998\end{array}$

Since, the probabilities of three specific sequence occurring randomly are similar, so it’s appropriate for everyone to choose sequence 2 as one of the most likely real results.

The probability that a coin will result in all tails or all heads is extremely small since there is only one sequence that result in A or B

Therefore, a sequence containing a combination of heads and tails is more likely to occurs than a sequence with all heads or all tails.