Answer:
If all you care about is whether you roll 2 or not, you get a Binomial distribution with an individual success probability 1/6. The probability of rolling 2 at least two times, is the same as the probability of not rolling 2 at zero or one time.
the answer is, 1 - bin(k=0, n=4, r=1/6) - bin(k=1, n=4, r=1/6). This evaluates to about 13%, just like your result (you just computed all three outcomes satisfying the proposition rather than the two that didn’t).
Step-by-step explanation:
Answer:
{HH, HT, TH, TT}
Step-by-step explanation:
The set of all possible outcomes in tossing a coin twice is;
{HH, HT, TH, TT}
In the first toss the coin may land Heads. In the second toss the coin may land Heads or Tails. This can be represented as;
HH, HT
Heads in the first and second tosses. Heads in the first toss followed by a Tail in the second toss.
In the first toss the coin is also likely to land Tails. In the second toss the coin may land Heads or Tails. This can be represented as;
TH, TT
Tails in the first toss followed by a Head in the second toss. Tails in the first and second tosses.
Combining these two possibilities will give us the set of all possible outcomes in tossing a coin twice is;
{HH, HT, TH, TT}
An exponential behavior can be observed.
Let x be the number of years she has to wait.
100((1+0.08/12)^12x) = 150
Now solve for x.
log base 1.08/12 of 1.5 = 61.02 approx
12x=61.02
x=5.09 years
Answer:
c
Step-by-step explanation:
i think it's 1
Answer:
Step-by-step explanation:
(X+3)(x-2)
3x-2x
X
