Answer:
Step-by-step explanation:
3 envelopes having 2 red card
2 envelopes having 1 red card and 1 black card
1 envelope having 2 black cards
We are given that . An envelope is selected at random and a card is withdrawn and found to be red.
So, No. of ways of envelope having red card = 3+2 = 5
No. of required ways of envelope having 1 red card and 1 black card = 2
So, probability of getting an envelope having 1 red card and 1 black card =
Hence The chance the other card is black is
The answer is 5. The only factors are 1 and 5, and 5 is the highest.
Answer:
Negative karma: You yell at someone for being dumb, but you fail your exam.
Good karma: You give money to the homeless, and then find money on the floor.
Answer:
<h2>____</h2><h3>1260 </h3><h2>____</h2>
Since we have to predict, we can use ratios to help us solve this.
12 said yes out of 72 randomly chosen, SO,
if 210 would say yes (goes to play), how many is total??
12 is to 72 AS 210 is to HOW MUCH (let it be x)??
We can setup ratio, cross multiply, and solve for x:
= 12/72= 210/x
= 12x = 210*72
=12x = 15,120
=x = 15,120/12
<h2>=x =1260</h2>
Step-by-step explanation:
the probability is always the number of desired cases over the number of all possible cases.
in our situation we have 15 cards.
that is the total possible cases when a random card is chosen.
how many desired cases do we have ?
a number NOT a multiple of 5.
how many are there ?
it is easier to say how many numbers there are being a multiple of 5 : 5, 10, 15
so, 3 numbers out of the 15 are multiple of 5.
that means
15 - 3 = 12 numbers of the 15 are NOT multiples of 5.
so, the probability to draw a card that is not a multiple of 5 is
12/15 = 4/5 = 0.8
the information about event B and even numbers is irrelevant for the question.
