Abag contains $3$ white chips and $3$ red chips. you repeatedly draw a chip at random from the bag. if it's white, you set it aside; if it's red, you put it back in the bag. after removing all $3$ white chips, you stop. what is the expected number of times you will draw from the bag?

total number of white chips=3

total number of red chips=3

total number of chips in the bag= 3 w +3 r=6

now consider expectancy of getting first white ball

pr(white chip) = 1/2

pr(red chip) = 1/2

now suppose while drawing chips from the bag in starting white chip packet comes out, or there may be chance that in second draw you are getting white chip packet, or in third

so, e(x)=1×1/2+2×1/2×1/2+3×(1/2)²×1/2+4×(1/2)³×1/   )

multiplying both sides by 1/2,the equation reduces to

1/2×e(x)=           1×(1/2)²+2×(1/2)³+

e(x)/2=1

e(x)=2

so the expected value of getting first white chip packet = 2

now when first white chip packet is drawn there are 2 white and 3 red chips packet.

p(w )=2/5,   p(r )=3/5

let e(v) denote the expectancy of drawing second white ball.

e(v) =1×2/5 +2×3/5×2/5+3×[3/5]²×2/5+4×[3/5]³×2/5+)

multiplying both sides by 3/5, we get

e(r)=4

total expectancy of getting 3 white ball=e(x) +e(v) +e(r)

                                                                  =2+5/2+4

                                                                    =4.5 +6

                                                                      =10.5

e[3 white chips] = 10.5

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.

lol we need to see what the question is asking about

step-by-step explanation: