Question

A fair coin is tossed n times. Let X = the number of heads obtained. If (X = 4), P(X = 5) and P(X = 6) are in A.P., then n is equal to
(a) 7
(b) 10
(c) 1
(d) 14

Answer 1

Hint : Here, we use the concept of binomial distribution, as heads or tails are only outcomes. Use formula of r successes for X = 4, X = 5 and X = 6. Also it is given successes are in A.P. so relate the tree using the mean of A.P. formula. By solving the equation obtained find the value of n.

Complete step-by-step answer :
When a coin is tossed, we get either head or tail.
Probability of getting is $p = \dfrac{1}{2}$
Probability of getting tail i.e., probability of not getting head as outcome = 1 – p = $= 1 - \dfrac{1}{2} = \dfrac{1}{2}$
Applying formula of r successes
$\therefore P(X = r) = {}^n{C_r}{p^r}{\left( {1 - p} \right)^{n - r}} = {}^n{C_r}{\left( {\dfrac{1}{2}} \right)^n}$
Since P(X = 4), P(X = 5) and P(X = 6) are in A.P.
Also it is given that; three terms are in A.P. then twice of mid-term is equal to sum of its extremes.
∴ 2P(X = 5) = P(X = 4) + P(X = 6)
$\Rightarrow 2 = \dfrac{{P(X = 4)}}{{P(X = 5)}} + \dfrac{{P(X = 6)}}{{P(X = 5)}}$
Putting values
$\Rightarrow 2 = \dfrac{{{}^n{C_4}}}{{{}^n{C_5}}} + \dfrac{{{}^n{C_6}}}{{{}^n{C_5}}}$
Simplifying the equation
$\Rightarrow 2 = \dfrac{{n!}}{{4!(n - 4)!}} \times \dfrac{{5!(n - 5)!}}{{n!}} + \dfrac{{n!}}{{4!(n - 4)!}} \times \dfrac{{5!(n - 5)!}}{{n!}}$
Simplifying values and cancelling factorial values
$\Rightarrow 2 = \dfrac{5}{{n - 4}} + \dfrac{{n - 5}}{6}$
Rearranging the using and represented in quadratic equation form.
⇒ ${n^2} - 21n + 98 = 0$
Solving quadratic equation, using factorization method
$\Rightarrow {n^2} - 14n - 7n + 98 = 0$
⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 14) (n – 7) = 0
⇒ n – 14 = 0 OR n – 7 = 0
⇒ n = 14 OR n = 7

So, the correct answer is “Option A AND D”.

Note : In these types of questions, use the formula of binomial distribution, so must understand what type of probability is asked. Also use the concept of A.P. as this is given in question. Solve the equation very carefully because in the case of binomial distribution you will always get complex equations as it involves factorial. Always try to cancel factorial terms by one another, do not find the value of factorial.