**K: Number of tosses till the first head appears **

$K=p\times 1+ (1-p)\times (K+1)$

$K=p+K+1-Kp-p$

$K=\dfrac{1}{p}$

$Ans: A$

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35 votes

When a coin is tossed, the probability of getting a Head is $p, 0 < p < 1$. Let $N$ be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of $N$ is

- $\dfrac{1}{p}$
- $\dfrac{1}{(1 - p)}$
- $\dfrac{1}{p^{2}}$
- $\dfrac{1}{(1 - p^{2})}$

9

1.If we get a tail in the first throw that means we have wasted one throw..ie (probability of getting tail)(N+1)= (1-p)(N+1)

2.If we get a head on the first throw than we are done.. ie (Probability of getting heads)(No. of tosses)=(p)(1)=p

Summing both up..we get.. N=(1-p)(N+1)+ p

After solving we get N=1/p

2.If we get a head on the first throw than we are done.. ie (Probability of getting heads)(No. of tosses)=(p)(1)=p

Summing both up..we get.. N=(1-p)(N+1)+ p

After solving we get N=1/p

2

45 votes

Best answer

$E = 1 \times p + 2 \times (1 - p)p + 3 \times (1 - p)(1 - p)p + \dots$

multiply both side with $(1 - p)$ and subtract:

$E - (1 - p)E = 1 \times p + (1 - p)p + (1 - p)(1 - p)p + \dots$

$\quad = p /(1 - (1 -p)) = 1$ (because it is now forming a GP)

$\quad \implies (1 - 1 + p)E = 1$

$\quad \implies E = 1 / p$

So, option $(A).$

multiply both side with $(1 - p)$ and subtract:

$E - (1 - p)E = 1 \times p + (1 - p)p + (1 - p)(1 - p)p + \dots$

$\quad = p /(1 - (1 -p)) = 1$ (because it is now forming a GP)

$\quad \implies (1 - 1 + p)E = 1$

$\quad \implies E = 1 / p$

So, option $(A).$

edited
Jan 15, 2019
by araj52270

**N be the random variable denoting the number of tosses till the first Head appears,**

**Toss1 "tail" * toss2 "tail"*............upto N-1 times * head appears only one time in last**

**{(1-p)^N-1] * p**

**is it right??????????**

0

25 votes

Let X be the number of tosses of a fair coin until a head appears, and we want to find E(X). When we toss the coin once, there are two possibilities:

- a
**head**occurs with a probability*p*. - a
**tail**occurs with a probability*(1-p)*. In this case the expected number of trials would be (1 + E(X)) where 1 is from the lost trial

∴ E(X) = p*1 + (1-p)*(1+E(X))

Solving it gives us E(X)=1/p.

9 votes

Tosses | Probability |

1 | p |

2 | (1-p)p |

3 | (1-p)(1-p)(p) |

4 | (1-p)(1-p)(1-p)p |

. | . |

. | . |

. | . |

. | . |

n | $(1-p)^{n-1} \times p$ |

$E = 1 \times p + 2 \times p(1-p) + 3 \times (1-p)^{2}p + ......... $

$E = \Large \sum_{n=1}^{\infty} n \times (1-p)^{n-1} \times p $

$E = \Large p \sum_{n=1}^{\infty} n \times (1-p)^{n-1} $

$E = \Large p \sum_{n=0}^{\infty} (n+1) \times (1-p)^{n} $

and we know that,

$\Large \sum_{k=0}^{\infty} (k+1)x^k = 1 + 2x + 3x^2 + 4x^3+..... = \frac{1}{(1-x)^2}$

Hence,

$E = \frac{1}{(1-(1-p))^2} \times p$

$E = \frac{1}{(p)^2} \times p$

$E \large = \frac{1}{p}$

7 votes

Actually, this is Geometric distribution.Here N is Geometric r.v which represents no of tosses required for first head.

The expected value of a Geom r.v N = **1/P**

P(Head) = p

So,

**E(N) = 1/p**

edited
Oct 1, 2019
by Gaurav Yadav

For reference, one can see what is expected value of geometric random variable here:

Expected Value of Geometric Random Variable..Proof @ Khan Academy

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