Question

If a fair coin is tossed 5 times, what is the probability that heads does not occur two or more times in a row?
(a). $\dfrac{{12}}{{{2^5}}}$
(b). $\dfrac{{13}}{{{2^5}}}$
(c). $\dfrac{{14}}{{{2^5}}}$
(d). $\dfrac{{15}}{{{2^5}}}$

Answer 1

Hint- We will just approach this question in such a way that we will find the general formula for this by just taking into consideration strings consisting of H, T of length 1, 2, 3 and so on in which the heads does not occur two or more times in a row. In this way we will just calculate the no. of possible strings of length 5 in which heads does not occur two or more times in a row.

Complete step-by-step answer:

Let ${a_n}$ be the number of strings of H and T of length n with no two adjacent H’s.
Let us enlist the base cases for the strings.
So, ${a_1} = 2$ (either H or T)
Also, ${a_2} = 3$ (either HT, TH or TT)
Now, we will just come into conclusion of a general case for ${a_{n + 2}}$
We get to know that ${a_{n + 2}} = {a_{n + 1}} + {a_n}$ , because the string must begin with either T or HT.
So, on substituting n = 1 we get ${a_{1 + 2}} = {a_{1 + 1}} + {a_1}$
$\Rightarrow {a_3} = {a_2} + {a_1}$
On substituting the value of ${a_2},{a_1}$ we get
$\Rightarrow {a_3} = 3 + 2$
$\Rightarrow {a_3} = 5$
So, on substituting n = 2 we get ${a_{2 + 2}} = {a_{2 + 1}} + {a_2}$
$\Rightarrow {a_4} = {a_3} + {a_2}$
On substituting the value of ${a_3},{a_2}$ we get
$\Rightarrow {a_4} = 5 + 3$
$\Rightarrow {a_4} = 8$
So, on substituting n = 3 we get ${a_{3 + 2}} = {a_{3 + 1}} + {a_3}$
$\Rightarrow {a_5} = {a_4} + {a_3}$
On substituting the value of ${a_4},{a_3}$ we get
$\Rightarrow {a_5} = 8 + 5$
$\Rightarrow {a_5} = 13$
Now, we have found that ${a_5} = 13$ which simply represents the number of strings of length 5 (because the fair coin is tossed for the 5 times) consisting of H, T in which heads does not occur two or more times in a row.
Now, we know that when a fair coin is tossed for the 5 times, then the total number of possible combinations so formed $= {2^5}$
Let the Number of outcomes favorable to an event E be p and the total number of outcomes be q
Probability of the event (E) = $\dfrac{p}{q}$
Now, probability that heads does not occur two or more times in a row is simply number of strings of length 5 (because the fair coin is tossed for the 5 times) consisting of H, T in which heads does not occur two or more times in a row divided by the number of total combinations so forms when a fair coin is tossed 5 times i.e. $\dfrac{{13}}{{{2^5}}}$
Hence, probability that heads does not occur two or more times in a row is $\dfrac{{13}}{{{2^5}}}$
$\therefore$ Option B. $\dfrac{{13}}{{{2^5}}}$ is the correct answer.

Note- This question can be solved using a manual approach in such a way that we will just write down all the possible combinations of strings of length 5 consisting of H, T so formed one by one when a fair coin is tossed 5 times. The possible combinations are HTTTT, HTHTT, HTHTH, TTTTT, THTTT, TTHTT, TTTHT, TTTTH, HTTHT, HTTTH, THTHT, THTTH and TTHTH i.e. total of 13 combinations. Now, total possible combinations when a fair coin is thrown 5 times is ${2^5}$ . Hence, the probability that heads does not occur two or more times in a row is $\dfrac{{13}}{{{2^5}}}$ .