Binomial distribution Q (1 Viewer)

mr.habibbi · Aug 30, 2021

Hi all,

Can someone help spot my error? Answer is n=17 apparently.

Q: A card is selected from a pack, its suit is noted, and it is returned. How many times must this be done so that the probability of at least one heart is more than 99%?

$P(X>=1) = 1-P(X=0) > 0.99$

P(X=0) < 0.01

solve for n:
$p(no hearts) = 3/4$
$q(hearts) = 1/4$
$P(X=0) = nC0(3/4)^0(1/4)^{n-0} < 0.01$
$(1/4)^n < 0.01$
n > 3.32, hence n = 4 as neZ+

notme123 · Aug 30, 2021

You did the binomial expansion wrong.
it should be the (1/4)^0 since the heart (1/4 prob) is picked up 0 times. its (3/4)^n < 0.01
You get n=16.0078, hence n = 17

mr.habibbi · Aug 30, 2021

notme123 said:
You did the binomial expansion wrong.
it should be the (1/4)^0 since the heart (1/4 prob) is picked up 0 times. its (3/4)^n < 0.01
You get n=16.0078, hence n = 17

success = 1/4 only if we're solving for P(X>=1) but ive rearranged it to P(X=0), so success is 3/4 ?

CM_Tutor · Sep 1, 2021

$\begin{align*} \text{We know that}\ P(\text{success}) &= P(\text{Heart}) = \cfrac{1}{4} \\ \text{And that}\ P(\text{failure}) &= P(\text{not a Heart}) = \cfrac{3}{4} \\ \text{From which it follows that}\ P(\text{$r$ Hearts from $n$ tries}) &= \binom{n}{r}\left(\cfrac{1}{4}\right)^r\left(\cfrac{3}{4}\right)^{n-r} \qquad \text{. . . . . . . . . . (*)} \end{align*}$

$\begin{align*} \text{Now,}\ P(\text{at least 1 Heart}) &= P(\text{1 Heart or 2 Hearts or 3 Hearts or ... or $n$ Hearts}) \\ &= P(\text{1 Heart}) + P(\text{2 Hearts}) + P(\text{3 Hearts}) + ... + P(\text{$n$ Hearts}) \\ &= \bigg[P(\text{0 Hearts}) + P(\text{1 Heart}) + P(\text{2 Hearts}) + P(\text{3 Hearts}) + ... + P(\text{$n$ Hearts})\bigg] - P(\text{0 Hearts}) \\ &= 1 - P(\text{0 Hearts}) \qquad \qquad \qquad \text{as there must be 0 or 1 or 2 or ... or $n$ Hearts in $n$ tries} \\ &= 1 - \binom{n}{0}\left(\cfrac{1}{4}\right)^0\left(\cfrac{3}{4}\right)^{n-0} \qquad \text{using (*)} \\ &= 1 - 1 \times 1 \times \left(\cfrac{3}{4}\right)^n \\ \text{So,}\ P(\text{at least 1 Heart}) &= 1 - \left(\cfrac{3}{4}\right)^n \end{align*}$

$\begin{align*} \text{And, we seek}\ P(\text{at least 1 Heart}) &> 99\% \\ 1 - \left(\cfrac{3}{4}\right)^n &> 0.99 \\ 0.01 &> 0.75^n \\ \ln{0.01} &> \ln{0.75^n} = n\ln{0.75} \\ \cfrac{\ln{0.01}}{\ln{0.75}} &< n \qquad \qquad \text{as $ln{0.75} < 0$} \\ 16.0078... &< n \\ \text{So,}\ n &\geqslant 17 \qquad \qquad \text{as $n \in \mathbb{Z}$} \end{align*}$

The minimum number of trials needed to be more than 99% certain that there has been at least one heart is 17, and the probability of at least one heart from $n$ trials will remain greater than 99% for all integers $n > 17$ .

Binomial distribution Q (1 Viewer)

mr.habibbi

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