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

Question

Travis Hicks · Accepted Answer

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

n = 14

and probability of success

p = 0.5

(Since,

P (T) = \frac{1}{2} = 0.5

)
Probability mass function of Binomial variable X is given by the formula:

P (X = x) =^{n} C_{x} p^{x} {(1 - p)}^{n - x}; x = 0, 1, 2 \dots . .14

Step 2
Required probability is obtained as follows:

P (X \geq 4) = 1 - P (X < 4)

= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)]

= 1 - [^{{14}} C_{0} {(0.5)}^{0} {(1 - 0.5)}^{14 - 0} +^{14} C_{1} {(0.5)}^{1} {(1 - 0.5)}^{14 - 1} +^{14} C_{2} {(0.5)}^{2} {(1 - 0.5)}^{14 - 2} +^{14} C_{3} {(0.5)}^{3} {(1 - 0.5)}^{14 - 3}]

= 1 - [\frac{14!}{0! (14 - 0)!} \times 1 \times 0.00006103515625 + \frac{14!}{1! (14 - 1)!} \times 0.5 \times 0.0001220703125 + \frac{14!}{2! (14 - 2)!} \times 0.25 \times 0.000244140625 + \frac{14!}{3! (14 - 3)!} \times 0.125 \times 0.00048828125]

= 1 - [1 \times 0.00006103515625 + 14 \times 0.00006103515625 + 91 \times 0.00006103515625 + 364 \times 0.00006103515625]

= 1 - [0.00006103515625 + 0.0008544921875 + 0.005554199219 + 0.02221679688]

= 1 - 0.02868652344

= 0.9713134766

\approx 0.9713

Thus, the probability that you achieve at least 4 tails is 0.9713

peterpan7117i · Accepted Answer

Step 1  Binomial distribution:  P(x)=nCx pxqn−x  P (tails), p=0.5  q=1−0.5=0.5  Number of trials, n=14  Pat least 4 tails}=1−P(less than 4 tails)  =1−[P(0)+P(1)+P(2)+P(3)]  =1−[0.514+14×0.5×0.513+14C2×0.52×0.512+14C3×0.53×0.511]  =0.9713

nick1337 · Accepted Answer

Probability of flipping either a head or a tail =12 Probability of flipping 6 tails in a row (12)6 Probability of flipping at least 1 head =1−(12)6=6364

Suppose that you flip a coin 14 times. What is

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