# Suppose that you flip a coin 14 times. What is

Suppose that you flip a coin 14 times. What is the probability that you achieve at least 4 tails?

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Travis Hicks
Step 1
Given Information:
Suppose that you flip a coin 14 times.
To find the probability that you achieve at least 4 tails:
Let X be a random variable which denotes the number of tails and X follows Binomial distribution with number of trials $$\displaystyle{n}={14}$$ and probability of success $$\displaystyle{p}={0.5}$$
(Since, $$\displaystyle{P}{\left({T}\right)}={\frac{{{1}}}{{{2}}}}={0.5}$$)
Probability mass function of Binomial variable X is given by the formula:
$$\displaystyle{P}{\left({X}={x}\right)}=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{n}-{x}}};{x}={0},{1},{2}\ldots.{.14}$$
Step 2
Required probability is obtained as follows:
$$\displaystyle{P}{\left({X}\geq{4}\right)}={1}-{P}{\left({X}{ < }{4}\right)}$$
$$\displaystyle={1}-{\left[{P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}+{P}{\left({X}={2}\right)}+{P}{\left({X}={3}\right)}\right]}$$
$$\displaystyle={1}-{\left[^{\left\lbrace{14}\right\rbrace}{C}_{{{0}}}{\left({0.5}\right)}^{{{0}}}{\left({1}-{0.5}\right)}^{{{14}-{0}}}+^{{{14}}}{C}_{{{1}}}{\left({0.5}\right)}^{{{1}}}{\left({1}-{0.5}\right)}^{{{14}-{1}}}+^{{{14}}}{C}_{{{2}}}{\left({0.5}\right)}^{{{2}}}{\left({1}-{0.5}\right)}^{{{14}-{2}}}+^{{{14}}}{C}_{{{3}}}{\left({0.5}\right)}^{{{3}}}{\left({1}-{0.5}\right)}^{{{14}-{3}}}\right]}$$
$$\displaystyle={1}-{\left[{\frac{{{14}!}}{{{0}!{\left({14}-{0}\right)}!}}}\times{1}\times{0.00006103515625}+{\frac{{{14}!}}{{{1}!{\left({14}-{1}\right)}!}}}\times{0.5}\times{0.0001220703125}+{\frac{{{14}!}}{{{2}!{\left({14}-{2}\right)}!}}}\times{0.25}\times{0.000244140625}+{\frac{{{14}!}}{{{3}!{\left({14}-{3}\right)}!}}}\times{0.125}\times{0.00048828125}\right]}$$
$$\displaystyle={1}-{\left[{1}\times{0.00006103515625}+{14}\times{0.00006103515625}+{91}\times{0.00006103515625}+{364}\times{0.00006103515625}\right]}$$
$$\displaystyle={1}-{\left[{0.00006103515625}+{0.0008544921875}+{0.005554199219}+{0.02221679688}\right]}$$
$$\displaystyle={1}-{0.02868652344}$$
$$\displaystyle={0.9713134766}$$
$$\displaystyle\approx{0.9713}$$
Thus, the probability that you achieve at least 4 tails is 0.9713
###### Not exactly what you’re looking for?
peterpan7117i
Step 1
Binomial distribution:
$$\displaystyle{P}{\left({x}\right)}={n}{C}{x}\ {p}^{{{x}}}{q}^{{{n}-{x}}}$$
P (tails), $$\displaystyle{p}={0.5}$$
$$\displaystyle{q}={1}-{0.5}={0.5}$$
Number of trials, $$\displaystyle{n}={14}$$
$$\displaystyle{P}\text{at least 4 tails}\rbrace={1}-{P}\text{(less than 4 tails)}$$
$$\displaystyle={1}-{\left[{P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}+{P}{\left({3}\right)}\right]}$$
$$\displaystyle={1}-{\left[{0.5}^{{{14}}}+{14}\times{0.5}\times{0.5}^{{{13}}}+^{{{14}}}{C}_{{{2}}}\times{0.5}^{{{2}}}\times{0.5}^{{{12}}}+^{{{14}}}{C}_{{{3}}}\times{0.5}^{{{3}}}\times{0.5}^{{{11}}}\right]}$$
$$\displaystyle={0.9713}$$
nick1337

Probability of flipping either a head or a tail
$$=\frac{1}{2}$$
Probability of flipping 6 tails in a row
$$(\frac{1}{2})^{6}$$
Probability of flipping at least 1 head
$$=1-(\frac{1}{2})^{6}=\frac{63}{64}$$