Answer: The correct options are
(1) (D) 0.30
(2) (A) 20%
(3) (C) 24.5%.
Step-by-step explanation: We are given to answer all the following three questions.
(1) Given that A and B are independent events, where
[tex]P(A)=0.80,~~P(A\cap B)=0.24,~~~P(B)=?[/tex]
We know that
if S and T are independent events, then
[tex]P(S\cap T)=P(S)\times P(T).[/tex]
Therefore, we get
[tex]P(A\cap B)=P(A)\cap P(B)\\\\\Rightarrow 0.24=0.80\times P(B)\\\\\Rightarrow P(B)=\dfrac{0.24}{0.80}\\\\\Rightarrow P(B)=0.30.[/tex]
Option (D) is CORRECT.
(2) Given that a weather forecaster predicts that their is 50% chance of rain on Saturday and a 40% chance of rain on Sunday.
We are to find the probability that it will rain both days.
Let X and Y represents the probabilities that it will rain on Saturday and Sunday respectively.
Then, we have
[tex]P(X)=50\%=\dfrac{50}{100}=\dfrac{1}{2},\\\\\\P(Y)=40\%=\dfrac{40}{100}=\dfrac{2}{5}.[/tex]
Since X and Y are independent of each other, so the probability that it will rain both days is
[tex]P(X\cap Y)=P(X)\times P(Y)=\dfrac{1}{2}\times\dfrac{2}{5}=\dfrac{1}{5}\times100\%=20\%.[/tex]
Option (A) is CORRECT.
(3) Given that a card is randomly drawn from a shuffled deck of cards and NOT REPLACED. A second card is drawn from the remaining shuffled cards.
We are to find the probability that both cards are RED.
Since there are 26 red cards in a pack of 52 cards, so the probability of drawing first red card is
[tex]p_1=\dfrac{26}{52}=\dfrac{1}{2}.[/tex]
Without replacement, the probability of drawing second red card will be
[tex]p_2=\dfrac{25}{51}.[/tex]
Therefore, the probability that both cards are red is
[tex]p=p_1\times p_2=\dfrac{1}{2}\times\dfrac{25}{51}=\dfrac{25}{102}=0.245\times100\%=24.5\%.[/tex]
Option (C) is CORRECT.
Thus, (D), (A) and (C) are correct options.
