Question

a biased coin is tossed thrice. what is the probability ythat tail turns out at least twice considering that the probability of a head is 60%

Answer 1

This is a binomial distribution.

Bin (n, p) = ⁿₓPˣ (1 - P)ⁿ⁻ˣ

n = number of tosses

p = probability of getting a head

Doing the substitution we have :

Bin (3,0.6) = ³x(0.6)ˣ (1 - 0.6)³⁻ˣ

x = 0,1, 2, 3

The probability that a tail is obtained at least twice.

= P( x = 0) or P(x = 1)

Doing the substitution we have :

³₀(0.6)⁰ (0.4)³⁻⁰ + ³₁(0.6) (0.4)³⁻¹

= 3!/(3 - 0)! 0! × 0.6 ⁰× 0.4³ = 0.064

= 3!/(3 - 1)! 1! × 0.6 × 0.4² = 0.288

Adding the two :

0.064 + 0.288 = 0.352

The answer is thus : 0.352

Answer 2

$\textbf {Hey there, here is the solution.}$
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P(head) = 60% = 0.6 (Given)

Therefore, P(tail) = 100 - 60 = 40% = 0.4

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Find the probability of at least 2 tails:

P( tail, tail, head) = 0.4 x 0.4 x 0.6 = 0.096

P(tail, head, tail) = 0.4 x 0.6 x 0.4 = 0.096

P(head, tail, tail) = 0.6 x 0.4 x 0.4 = 0.096

P(tail, tail, tail) = 0.4 x 0.4 x 0.4 = 0.064

P(at least 2 tails) = 0.096 + 0.096 + 0.096 + 0.064 = 0.352

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Answer: The probability of getting at least 2 tails is 0.352
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⭐$\textbf {Cheers}$⭐