Answer:
Check the explanation
Step-by-step explanation:
let the money on bet is X.
probability of winning =18/38=9/19
probability of losing =(1-9/19)=10/19
expected outcome =
probability *return =(
Expected value of return after one bet is =(9/19*x)-(10/19*x)=-1x/19
it is negative which is obvious cause casinos are there to earn money.
a) Our best strategy in this case as probability of winning is near by 50 %, we should place a bet of 1 $ each,and when we lose one bet consecutively we should bet twice the amount..
Cause two consecutive losses on black has less probability.
c) In case we have to reach 30 $ we have to use the same strategy as above.
Answer: 0.0241
Step-by-step explanation:
This is solved using the probability distribution formula for random variables where the combination formula for selection is used to determine the probability of these random variables occurring. This formula is denoted by:
P(X=r) = nCr × p^r × q^n-r
Where:
n = number of sampled variable which in this case = 21
r = variable outcome being determined which in this case = 5
p = probability of success of the variable which in this case = 0.31
q= 1- p = 1 - 0.31 = 0.69
P(X=5) = 21C5 × 0.31^5 × 0.69^16
P(X=5) = 0.0241
#25.
Convert the fractional numbers into decimals for easier comparison.
3 2/3 is approximately 3.66
-4 2/5 is approximately -4.4
We now have: 3, 3.66, -4.2, -4.4
In order: -4 2/5, -4.2, 3, 3 2/3
#26.
All positive numbers are greater than negative numbers.
When comparing two negative numbers, the one with the larger absolute value, will be less.
So (G) is correct
#27.
25 = m + 6.3
25 - 6.3 = m + 6.3 - 6.3
m = 18.7
#28.
h = 3.2-1.6 = 1.6
#29.
x = 15.6 - 9.8 = 5.8
#30
p = 17 - 4.5 = 12.5
Let (a) be the first cat
let (b) be the second cat
a +b=11kg-----(1)
a=b+1500g⇒(1.5 kg)-------(2)
so now u have a system of two unknowns
a=11-b
substitute a in (2)
11-b=b+1.5
11-1.5=2b
2b=9.5
b=9.5/2=4.75
subs. b in (1)
a+b=11
a+4.75=11
a=11-4.75=6.25
so cat (a) is 6.25 kg
and (b) is 4.75 kg
hope I helped
