Suppose that you flip a coin 14 times. What is

Wanda Kane 2021-12-21 Answered
Suppose that you flip a coin 14 times. What is the probability that you achieve at least 4 tails?

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Expert Answer

Travis Hicks
Answered 2021-12-22 Author has 1811 answers
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
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peterpan7117i
Answered 2021-12-23 Author has 2762 answers
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, \(\displaystyle{n}={14}\)
\(\displaystyle{P}\text{at least 4 tails}\rbrace={1}-{P}\text{(less than 4 tails)}\)
\(\displaystyle={1}-{\left[{P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}+{P}{\left({3}\right)}\right]}\)
\(\displaystyle={1}-{\left[{0.5}^{{{14}}}+{14}\times{0.5}\times{0.5}^{{{13}}}+^{{{14}}}{C}_{{{2}}}\times{0.5}^{{{2}}}\times{0.5}^{{{12}}}+^{{{14}}}{C}_{{{3}}}\times{0.5}^{{{3}}}\times{0.5}^{{{11}}}\right]}\)
\(\displaystyle={0.9713}\)
0
nick1337
Answered 2021-12-28 Author has 10160 answers

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
\(=1-(\frac{1}{2})^{6}=\frac{63}{64}\)

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