Hey Guys, i am stuck understanding the word play in these two questions:
1. Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?
A. 5/7
B. 23/24
C. 6/7
D. 34/35
E. 8/13
2. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?
A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 11/32
Can someone help me understand how are these two questions different in terms of what has been asked.
What is the difference b/w "all not blue" and "both aren't blue out of two" ?
And why the logic fits for the 1st question and not the 2nd one. P(no blue)=1- (all blue)
Answer to 1st question can be calculated in both ways:
Method 1: (3C1*4C2)/7C3 + (3C2*4C1)/7C3 + 4C3/7C3 = 18/35 + 12/35 + 4/35 = 34/35.
Method 2: 1- 3C3/7C3 = 34/35
But for 2nd question: we can't use the 1 - P method:
Method 1: (First card is not blue and second is not blue) 6/8 * 5/7 = 15/28
Method 2: 1-2C2/8C2 = 27/28 (And this is incorrect)
1. Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?
A. 5/7
B. 23/24
C. 6/7
D. 34/35
E. 8/13
2. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?
A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 11/32
Can someone help me understand how are these two questions different in terms of what has been asked.
What is the difference b/w "all not blue" and "both aren't blue out of two" ?
And why the logic fits for the 1st question and not the 2nd one. P(no blue)=1- (all blue)
Answer to 1st question can be calculated in both ways:
Method 1: (3C1*4C2)/7C3 + (3C2*4C1)/7C3 + 4C3/7C3 = 18/35 + 12/35 + 4/35 = 34/35.
Method 2: 1- 3C3/7C3 = 34/35
But for 2nd question: we can't use the 1 - P method:
Method 1: (First card is not blue and second is not blue) 6/8 * 5/7 = 15/28
Method 2: 1-2C2/8C2 = 27/28 (And this is incorrect)