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lacy draws a heart from a standard deck of 52 cards. without replacing …

Question

lacy draws a heart from a standard deck of 52 cards. without replacing the first card, she then proceeds to draw a second card and gets a club.
are these events independent?
input yes or no:
determine the probability of drawing a heart and then a club without replacement.
write your answer in decimal form, rounded to four decimal places as needed.
answer =
linda draws a heart from a standard deck of 52 cards. she returns the heart to the deck, then draws a second card. her second card is a club.
are these events independent?
input yes or no:
determine the probability of drawing a heart and then a club with replacement.
write your answer in decimal form, rounded to four decimal places as needed.
answer =

Explanation:

Step1: Check independence for Lacy's case

Two events \(A\) (drawing a heart) and \(B\) (drawing a club) are independent if \(P(B|A)=P(B)\).
For Lacy, \(P(A)=\frac{13}{52}\), after drawing a heart (without replacement), \(P(B|A)=\frac{13}{51}\), and \(P(B)=\frac{13}{52}\). Since \(\frac{13}{51}
eq\frac{13}{52}\), the events are not independent.

Step2: Calculate probability for Lacy's case (without replacement)

Use the formula \(P(A\cap B)=P(A)\times P(B|A)\)
\(P(A)=\frac{13}{52}\), \(P(B|A)=\frac{13}{51}\)
\(P(A\cap B)=\frac{13}{52}\times\frac{13}{51}=\frac{169}{2652}\approx0.0637\)

Step3: Check independence for Linda's case

For Linda, since the card is replaced, \(P(A)=\frac{13}{52}\), \(P(B|A)=\frac{13}{52}\), and \(P(B)=\frac{13}{52}\). Since \(P(B|A) = P(B)\), the events are independent.

Step4: Calculate probability for Linda's case (with replacement)

Use the formula \(P(A\cap B)=P(A)\times P(B)\)
\(P(A)=\frac{13}{52}\), \(P(B)=\frac{13}{52}\)
\(P(A\cap B)=\frac{13}{52}\times\frac{13}{52}=\frac{169}{2704}= 0.0625\)

Answer:

For Lacy:

  • Input Yes or No: No
  • Answer = \(0.0637\)

For Linda:

  • Input Yes or No: Yes
  • Answer = \(0.0625\)