Box I contains 4 red and 8 blue marbles while box II contains 5 red and 3 blue marbles. An unfair coin in tossed, whose probability of turning up heads is 40%. If the coin comes up heads box I is chosen and a random marble is chosen, otherwise if it is tails the marble is chosen from box II.

(a) Find the probability a red marble is chosen.

(b) If a red marble is chosen, what is the probability it came from box I?

(c) If a blue marble is chosen, what is the probability it came from box I?

Suppose after the first marble is chosen, the experiment is repeated. Assume the first marble is NOT put back into its box. The coin is flipped again and another marble is chosen from either box I or box II.

(d) What is the probability that the second marble has the same color as the first marble?

(a) 0.5083

(b) 0.1333

(c) 0.2667

(d) 0.3596

Step-by-step explanation:

Box 1 = 4R and 8B

Box 2 = 5R and 3B

P(Box 1) = P(Head) = 40% = 0.4

P(Box 2) = P(Tail) = 1 - 0.4 = 0.6

(a) P(red marble) = (P(Box 1) and P(Red in box 1)) or (P(Box 2) and P(Red in box 2)) = (0.4 × 4/12) + (0.6 × 5/8) = 0.5083

(b) P(red marble from box 1) = (P(Box 1) and P(Red in box 1) = 0.4 × 4/12 = 0.1333

(c) P(blue marble from box 1) = (P(Box 1) and P(Blue in box 1) = 0.4 × 8/12 = 0.2667

(d) P(Two marbles with the same color) = P(1st Red from box 1 and 2nd Red from box 1) or P(1st Red from box 1 and 2nd Red from box 2) or P(1st Red from box 2 and 2nd Red from box 2) or P(1st Blue from box 1 and 2nd Blue from box 1) or P(1st Blue from box 1 and 2nd Blue from box 2) or P(1st Blue from box 2 and 2nd Blue from box 2)

P(1st Red from box 1 and 2nd Red from box 1) = (0.4 × 4/12) × (0.4 × 3/11) = 0.0145

(It is 11 in the second term because the first red has been removed.)

P(1st Red from box 1 and 2nd Red from box 2) = (0.4 × 4/12) × (0.6 × 5/8) = 0.0500

P(1st Red from box 2 and 2nd Red from box 2) = (0.6 × 5/8) × (0.6 × 4/7) = 0.1286

P(1st Blue from box 1 and 2nd Blue from box 1) = (0.4 × 8/12) × (0.4 × 7/11) = 0.0679

P(1st Blue from box 1 and 2nd Blue from box 2) = (0.4 × 8/12) × (0.6 × 3/8) = 0.0600

P(1st Blue from box 2 and 2nd Blue from box 2) = (0.6 × 3/8) × (0.6 × 2/7) = 0.0386

P(Two marbles with the same color) = 0.0145 + 0.05 + 0.1286 + 0.0679 + 0.06 + 0.0386 = 0.3596

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: