<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:
The correct answer is A) 4/663.
Step-by-step explanation:
First you find the probability of drawing a queen when drawing a single card from a deck of 52 cards. Since there are 4 queens(the queen of diamond, the queen of hearts, the queen of spades, and the queen of clubs) in a deck of 52 cards, the probability of drawing a queen when drawing a single card from a deck of 52 cards is 4/52.
Next you find the probability of drawing a king when drawing a single card from a deck of 51 cards(since you did not replace the first card you drew). Since there are 4 kings(the king of diamond, the king of hearts, the king of spades, and the king of clubs) in a deck of cards, the probability of drawing a king when drawing a single card from a deck of 51 cards is 4/51.
Then you multiply the two probabilities to determine the probability of drawing a queen then a king. So,
4/52 x 4/51 =
4 x 4/52 x 51 =
16/2652
Finally, simplify the fraction. The greatest number that can go into both the numerator and denominator is 4. So divide both the numerator and denominator by 4. When you do this, you get the following:
16 divided by 4 = 4 as the numerator and
2652 divided by 4 = 663 as denominator.
So, the final answer is 4/663.
We know that 12 equals one dozen. It is not asking for a dozen, but for 9 dozen. So, we need to multiply 9 by 12. 9*12 = 108.
It says the rolls of first aid tape were divided equally into 4 boxes. We need to divide these 108 rolls into 4 equal groups, since that is what took place in the problem. So, we divide 108 by 4.
108/4 = 27. There are 27 rolls in each box.
