Heads or tails

Probability Level 3

A hundred identical coins each with probability p p showing up heads are tossed once with 0 < p < 1 0<p<1 .

If the probability of heads showing up on 50 coins is equal to that of heads showing on 51 coins, find the value of p p .

Assume each toss is independent from each other.

59 100 \frac{59}{100} 49 101 \frac{49}{101} 1 2 \frac12 51 101 \frac{51}{101} 51 100 \frac{51}{100}

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Parth Tandon by parth tandon , 24, India

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2 solutions

Zubair Shaikh
May 15, 2015

By Binomial Probability Distribution

100 C 50 × p 50 × ( 1 − p ) 50 = 100 C 51 × p 51 × ( 1 − p ) 49 ^{100}C_{50} \times p^{50} \times (1-p)^{50} = ^{100}C_{51} \times p^{51} \times (1-p)^{49}

100 ! 50 ! × 50 ! × p 50 × ( 1 − p ) 50 = 100 ! 51 ! × 49 ! × p 51 × ( 1 − p ) 49 \frac{100!}{50!\times50!} \times p^{50} \times (1-p)^{50} = \frac{100!}{51!\times49!} \times p^{51} \times (1-p)^{49} ( 1 − p ) 50 ! × 50 × 49 ! = p 51 × 50 ! × 49 ! \frac{(1-p)}{50!\times50\times49!} = \frac{p}{51\times50!\times49!} ( 1 − p ) 50 = p 51 \frac{(1-p)}{50} = \frac{p}{51} 51 × ( 1 − p ) = 50 × p 51\times(1-p) = 50\times p 51 − 51 × p = 50 × p 51 - 51\times p = 50\times p 51 = 101 × p 51 = 101\times p p = 51 101 p = \boxed{\frac{51}{101}}

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This is a binomial distribution based problem

P ( 50 heads ) = 100 C 50 ( p ) 50 ( 1 − p ) 50 P(\text{50 heads}) = ^{100}C_{50}(p)^{50}(1-p)^{50}

P ( 51 heads ) = 100 C 51 ( p ) 51 ( 1 − p ) 49 P(\text{51 heads}) = ^{100}C_{51}(p)^{51}(1-p)^{49}

Since both are equal equate both.

You will get p = 51 101 p = \boxed{\frac{51}{101}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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