# 6% of male are color-blind. Find the probability that the first color-blind will be found among the first 5 men tested.

6% of male are color-blind. Suggest a male groub is randomly selected and tested, find the probability that the first color-blind will be found among the first 5 men tested.

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Rivka Thorpe

Geometric probability formula:
$$\displaystyle{P}{\left({X}={k}\right)}={q}^{{{k}-{1}}}{p}={\left({1}-{p}\right)}^{{{k}-{1}}}{p}$$
p=6%
k=1, 2, 3, 4, 5:
$$\displaystyle{P}{\left({X}={1}\right)}={\left({1}-{0.06}\right)}^{{{1}-{1}}}{\left({0.06}\right)}={0.94}^{{0}}{\left({0.06}\right)}={0.06}$$
$$\displaystyle{P}{\left({X}={1}\right)}={\left({1}-{0.06}\right)}^{{{2}-{1}}}{\left({0.06}\right)}={0.94}^{{1}}{\left({0.06}\right)}={0.0564}$$
$$\displaystyle{P}{\left({X}={1}\right)}={\left({1}-{0.06}\right)}^{{{3}-{1}}}{\left({0.06}\right)}={0.94}^{{2}}{\left({0.06}\right)}={0.053016}$$
$$\displaystyle{P}{\left({X}={1}\right)}={\left({1}-{0.06}\right)}^{{{4}-{1}}}{\left({0.06}\right)}={0.94}^{{3}}{\left({0.06}\right)}={0.049835}$$
$$\displaystyle{P}{\left({X}={1}\right)}={\left({1}-{0.06}\right)}^{{{5}-{1}}}{\left({0.06}\right)}={0.94}^{{4}}{\left({0.06}\right)}={0.0468449}$$
$$P(A\ or\ B)=P(A)+P(B)$$. So,
$$\displaystyle{P}{\left({X}\le{5}\right)}={P}{\left({X}={1}\right)}+{P}{\left({X}={2}\right)}+{P}{\left({X}={3}\right)}+{P}{\left({X}={4}\right)}+{P}{\left({X}={5}\right)}$$
=0.06+0.0564+0.053016+0.049835
=0.2660959
=26.60959%