Cards Conditional
A deck of 10 cards numbered 1 through 10 is shuffled, and a card is drawn from the deck. Let A be the event that an odd number is drawn and B be the event that a prime number is drawn. What is P ( B ∣ A ) ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Yes, I understand the concept, pls give me many qn through email adolphdamas55@gmail.com
But 1 is not considered to be a prime number now!
yes, exactly. One (1) is not considered to be a prime! (But B is the subset for primes in this example) https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/
Primes = { 1 ,2, 3 , 5 , 7 } = 5
Odds = { 1 , 3 , 5 , 7 ,9} = 5
Prime and Odd = { 1 , 3 , 5 , 7 } = 4
∣ T o t a l N u m b e r O f C a r d s ∣ ∣ P r i m e s ∣ + ∣ O d d s ∣ − ∣ P r i m e A n d O d d ∣ = 1 0 5 + 5 − 4 = 1 0 6 = 5 3
Is 1 a prime number?
I guess correct numerator is |PrimeAndOdd|.
1 is not a prime number - hence this solution is wrong ( https://uk.answers.yahoo.com/question/index?qid=20090904184713AA9yBs2&guccounter=1 )
What you're representing is not AND, that's exclusive OR = XOR A+B-(A&B).
Relevant wiki: Conditional Probability
A = { 1 , 3 , 5 , 7 , 9 }
B = { 2 , 3 , 5 , 7 }
A ∩ B = { 3 , 5 , 7 }
The sample space is uniform, so: P ( B ∣ A ) = ∣ A ∣ ∣ A ∩ B ∣ = 5 3