Practice33 When playing roulette at a casino, a gambler is trying to decide whether to bet on the number 33 or to bets that the outcome be any one of the five possibilities 00, 0, 1, 2 of 3. The gamble knows that the expected value of
the $5 bet for a single number is - 536. For the $S bet that the outcome is 00,0.12, or 3 there is a probability of te of making a net profit of $30 and a probability of losing 55
a. Find the expected value for the 35 bet that the outcome is 00, 0, 1, 2, 3
b. Which bet is better a 35 bet on the number 33 or a 55 bet that the outcome is any one of the numbers 00, 0.1.2013 Why?

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

the answer would be 0.54

4+3x hope you like mark as brainliest✔

c and d is your answers.

stay frosty!