You draw a card uniformly at random from a standard deck, then remove all cards of strictly higher rank (e. g., if you draw a 4, then the remaining deck consists of four 2s, four 3s, and three 4s). you repeat this process three times on the same deck, without putting the removed cards back. what is the probability that the three cards you’ve drawn are a 3 and two 5s?

≈ 0.0044

Step-by-step explanation:

The probability that the three cards you’ve drawn random from a standard  deck are a 3 and two 5s is possible only when you

first draw 5secondly draw  5thirdly draw 3

If one draws 3 first, then there is no probability that one draws 5 after, because all 5's are need to be removed after drawing 3.

The probability drawing 5 is . After removing cards with higher rank, remaining cards are 15. (four 2's, four 3's, four 4's and three 5's)

The probability of drawing another 5 is then . No card is removed else, since 5 is the highest rank. Thus, 14 cards left.

The probability of drawing 3 after two 5 is  

To find all these events happening successively, we need the multiply their probabilities:

× × =  ≈ 0.0044

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12