Question

One hundred identical coins, each with probability p, of showing up heads are tossed once. If $0$ < $p$ < $1$ and the probability of head showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is,
A) 1/2
B) 49/101
C) 50/101
D) 51/101

Answer 1

Hint: Here first we will use Bernoulli trial of probability in three of the cases to get the probability of success. After that we will compare the probability of heads showing on 50 coins and the same on 51 coins to get the value of p.

Complete step-by-step answer:
Let p = probability that coins shows head
1 – p = probability that coins does not show head
And n = number of coins
As per the Bernoulli trial of probability;
$P(r) = {}^n{C_r} \times {p^{^r}} \times {q^{n - r}}$
Where P(r) is the probability of r success in n trial
Here n is 100
Case-1
For 50 successes, r=50
Hence, $P(50) = {}^{100}{C_{50}} \times {p^{50}} \times {q^{100 - 50}}$
$ \Rightarrow P(50) = {}^{100}{C_{50}} \times {p^{50}} \times {(1 - p)^{100 - 50}}$
$ \Rightarrow P(50) = {}^{100}{C_{50}} \times {p^{50}} \times {(1 - p)^{50}}$………………………(1)
Case-2
For 51 success, r=51
Hence, $P(51) = {}^{100}{C_{51}} \times {p^{51}} \times {q^{100 - 51}}$
$ \Rightarrow P(51) = {}^{100}{C_{51}} \times {p^{51}} \times {(1 - p)^{100 - 51}}$
$ \Rightarrow P(51) = {}^{100}{C_{51}} \times {p^{51}} \times {(1 - p)^{49}}$…………………….(2)
As per the question, P(50) = P(51)………………..(3)
Putting equation 1 and 2 in equation 3 we get,
${}^{100}{C_{50}} \times {p^{50}} \times {(1 - p)^{50}} = {}^{100}{C_{51}} \times {p^{51}} \times {(1 - p)^{49}}$
Simplifying both the sides of the above equation we get,
${}^{100}{C_{50}} \times (1 - p) = {}^{100}{C_{51}} \times p$
Taking same terms to the same side we get,
$ \Rightarrow \dfrac{{1 - p}}{p} = \dfrac{{{}^{100}{C_{51}}}}{{{}^{100}{C_{50}}}}$
Putting formula of combination, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ on right hand side we get,
$ \Rightarrow \dfrac{{1 - p}}{p} = \dfrac{{\dfrac{{100!}}{{51! \times 49!}}}}{{\dfrac{{100!}}{{50! \times 50!}}}}$
Cancelling 100 from numerator and denominator of right hand side,
$ \Rightarrow \dfrac{{1 - p}}{p} = \dfrac{{50! \times 50!}}{{51! \times 49!}}$
Simplifying the above equation we get,
$ \Rightarrow \dfrac{{1 - p}}{p} = \dfrac{{50! \times 49! \times 50}}{{50! \times 51 \times 49!}}$
Cancelling identical term from numerator and denominator of right hand side,
$ \Rightarrow \dfrac{{1 - p}}{p} = \dfrac{{50}}{{51}}$
Simplifying again we get,
51(1-p) = 50p
Or, 51 – 51p = 50p
Taking -51p to the right hand side,
51= 101p
Or, p= 51/101
Hence the value of p is 51/101.
Option 4 is the correct answer.

Note: Be cautious while solving probability value and while comparing probability value of both the cases. You might make mistakes due to lack of concentration.
Combination is a mathematical technique that determines the number of possible arrangements in the collection of items where the order of selection does not matter. Its formula is, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Probability is concerned with numerical description of how likely an event is to occur and how likely it is that a proposition is true. It is always between 0 and 1.
Bernoulli trial of probability is a random experiment with exactly two possible outcomes i.e. success and failure.