6. An urn contains 15 red balls and 8 blue balls. In each draw, one ball is extracted at random. It is then returned to the urn, along with 6 extra balls of the same color. (the total number of balls in the urn increases after each draw). Consider the event Ck={a blue ball is extracted at the k-th draw}. Determine the probability P(C4). (at the first three draws the extracted balls are not blue).

P(C4) = 0.0711

Step-by-step explanation:

consider the first draw = 15/23  since it cannot be a blue ball

The second draw = 21/29 since 6 more red balls will be added after the draw since a blue ball cannot be drawn

the third draw = 27/35 since 6 more red balls will be added after each draw since a blue ball cannot be drawn

therefore the total number of red balls will be = 15 + 6 + 6 + 6 = 33 red balls after the 4th draw. the total ball now in the urn= 33 red + 4 blue = 41

Hence the probability of drawing a blue ball at the fourth draw after drawing red balls at the previous attempts = 8/41

P(C4) = P ( fourth ball is blue ) * P( first ball red)*P(second ball red) *P(third ball red )

= (8/41) * (15/23) * (21/29)* (27/35) = 0.0711

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

idk bring the function back   im not smart

step-by-step explanation: