1. two cards are drawn from a standard of 52 cards. Find the probability that the second card is a spade given that the first card is a club.
2. Suppose you draw 3 cards from a standard deck of 52 cards . Find the probability that the third card is a club given that the first two cards are spades.
3. AN urn contains 5 blue and 6 green balls. You take 2 balls out of the urn, one after the other, without replacement. Find the probability that the second ball is green given that the first ball is blue.
I need help with these questions. I do not know how to approach and answer them.This is under a subtopic conditional probability .
please help me.
Thanks in Advance.
Note that you are only asking for the probability of the last draw.
1. Assuming no replacement, there is one less card. 13/51
2. Use same reasoning.
3. Blue ball = 5/11, green ball = 6/(11-1)
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
Sure! I'd be happy to help you with these conditional probability questions.
Conditional probability is the probability of an event occurring given that another event has already occurred. In order to solve these problems, we need to understand the concept of conditional probability and apply the formula:
P(A | B) = P(A and B) / P(B)
where P(A | B) is the probability of event A happening given that event B has already occurred, P(A and B) is the probability of both A and B happening together, and P(B) is the probability of event B happening.
Now, let's solve the questions one by one:
1. The question asks for the probability that the second card is a spade given that the first card is a club.
There are 52 cards in total, and if the first card is a club, then there are 13 clubs left in the deck. Out of these remaining 51 cards, there are 13 spades.
Therefore, P(second card is a spade | first card is a club) = P(both cards are a club and a spade) / P(first card is a club)
= (13/52) / (13/52)
= 1/4
So, the probability that the second card is a spade given that the first card is a club is 1/4.
2. The question asks for the probability that the third card is a club given that the first two cards are spades.
If the first two cards are spades, then there are 50 cards remaining in the deck, out of which 11 are clubs.
Therefore, P(third card is a club | first two cards are spades) = P(both first two cards are spades and third card is a club) / P(first two cards are spades)
= (11/52) / (13/52)
= 11/13
So, the probability that the third card is a club given that the first two cards are spades is 11/13.
3. The question asks for the probability that the second ball is green given that the first ball is blue.
There are 11 balls in total, and if the first ball is blue, there are 5 blue balls and 6 green balls remaining.
Therefore, P(second ball is green | first ball is blue) = P(both balls are blue and green) / P(first ball is blue)
= (6/11) / (5/11)
= 6/5
So, the probability that the second ball is green given that the first ball is blue is 6/5.