Question

A fair cubical die is thrown four times. use the binomial probability formula to calculate the probability of at least two 3's.

Answer 1

By using the binomial probability formula, the probability of at least two 3's is equal to 0.1319.

How to calculate the probability of at least two 3's?

Since the fair cubical die is thrown four times, the number of times is given by n = 4 and the probability of 3's is P = 1/6.

Mathematically, the binomial probability formula is given by:

$P(X \geq x) =\sum^{n}_{r=x} ^nC_r (p)^r (q)^{(n-r)}\\\\P(X \geq 2) = \; ^4C_2 (\frac{1}{6} )^2 (\frac{5}{6} )^{(4-2)} + ^4C_3 (\frac{1}{6} )^3 (\frac{5}{6} )^{(4-3)} + ^4C_4 (\frac{1}{6} )^4 (\frac{5}{6} )^{(4-4)}\\\\P(X \geq 2) = 6 \times \frac{1}{36} \times \frac{25}{36} + 4 \times \frac{1}{216} \times \frac{5}{36} + 1 \times \frac{1}{1296} \times 1\\\\P(X \geq 2) = \frac{150}{1296} + \frac{20}{1296}+ \frac{1}{1296}$

P(X ≥ 2) = 19/144

P(X ≥ 2) = 0.1319.

Read more on probability here: brainly.com/question/25870256

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