Mathematics, 20.08.2020 17:01 muravyevaarina

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has faces numbered 1, 2, 3, 4, 4, and 4. A die is selected at random and rolled four times. You are told that two rolls were 1's and two were 4's. Find the probability the die chosen was green.

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Answer from: homelanie21

The probability the die chosen was green is 0.9

Step-by-step explanation:

From the information given :

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has faces numbered 1, 2, 3, 4, 4, and 4.

SO, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}= \dfrac{1}{2}$

A die is selected at random and rolled four times.

When the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

$= \begin {pmatrix} \left\begin{array}{c}4\\2 \end{array}\right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

$=\dfrac{4!}{2!(4-2)!}\times (\dfrac{1}{6})^4$

$=\dfrac{4!}{2!(2)!}\times (\dfrac{1}{6})^4$

$=\dfrac{4\times 3 \times 2!}{2!(2)!}\times (\dfrac{1}{6})^4$

$=\dfrac{12}{2 \times 1}\times (\dfrac{1}{6})^4$

$= 6 \times (\dfrac{1}{6})^4$

$= (\dfrac{1}{6})^3$

$= \dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

$= \begin {pmatrix} \left\begin{array}{c}4\\2 \end{array}\right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= \dfrac{4!}{2!(4-2)!} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= \dfrac{4!}{2!(2)!} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= 6 \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

$= \dfrac{9}{216}$

∴ The probability of two 1's and two 4's in both dies

= P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's = $(\dfrac{1}{216} \times \dfrac{1}{2} )+ ( \dfrac{9}{216} \times \dfrac{1}{2})$

The probability of two 1's and two 4's = $\dfrac{1}{432}+ \dfrac{1}{48}$

The probability of two 1's and two 4's = $\dfrac{5}{216}$

Using Bayes Theorem; the probability that the die was green can be computed as follows:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216} }{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{48} }{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{1}{48} \times \dfrac{216}{5 }$

P(second die (green) | two 1's and two 4's ) = $\dfrac{9}{10}$

P(second die (green) | two 1's and two 4's ) = 0.9

∴

The probability the die chosen was green is 0.9

Answer from: Quest

ddd

step-by-step explanation:

Idon't know why, it didn't show my pic last time, so here we go..

Answer from: Quest

step-by-step explanation:

one of the most important purposes of surface treatment is to alter surface properties according to the requirement of specific applications. for example, hydrophobic coating has been utilized on windshield and eyeglasses. gold atom monolayer on substrate has become a popular surface bioconjugation strategy.

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