Can you take a picture of the tables so i can see my choices??
<span>Winning Probablity = 0.2, hence Losing Probability = 0.8
Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0)
Winning once W1 is equal to L4, winning zero times is losing 5 times.
p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5);
p (W1) + p(W0) = p(L4) + p(L5)
p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5
p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>
Answer:
4
Step-by-step explanation:
If the letters are considered distinct, then the number of permutations is

.
If we count either C as the same character, then we would be double-counting - to correct this, we would simply divided by the number of ways we can choose C from the available characters, or

.
Answer:
the answer would be p= 15
Step-by-step explanation: