A bag contains 8 blue coins and 6 red coins. A coin is removed at random and replaced by three of the other color.
a) What is the probability that the removed coin is blue?
b) If the coin removed is blue, what is the probability of drawing a red coin after three red coins are put in the bag to replace the blue one?
c) If the coin removed is red, what is the probability of drawing a red coin after three blue coins are put in the bag to replace the red one?
THANK YOU!!!
Your all wrong! Becuase the answer changes every time you completely get it wrong!
bro just shut up its a new year and jusst gimme the answer aqlready
It is
4/7
9/16
5/16
You are all wrong it is 3/7.
ITS 9/16
its 4/17 people
3/10
The probability of drawing a red coin after three blue coins are put in the bag to replace the red one that has been removed is
8+3=11 6-1=5 11+5=16 5/16
9/16
A. You would add the number of coins there are (8+6=14) and then find what the question is asking for, which is the probability of removing a blue coin. The answer would be 8/14.
B. You would subtract one blue coin (8-1=7) and add three red ones. (6+3=9) You would find the probability the same way as A. :-)
your all wrong its
10/17
All of those people got it wrong!!
A) 4/7
B) 9/16
C) 4/9