One box has 7 red balls and 3 white balls; a second box has 6 red balls and 4 white balls. A pair of dice are tossed. If the sum of the dice is less than five, a ball is selected from the first box, otherwise the ball is selected from the second box. Find the probability of getting a red ball.
Answers
Answer:
(1/6)(7/10)+(5/6)(3/5)=37/60=0.617
4) Bayes Formula: One urn has 4 red balls and 1 white ball; a second urn has 2 red balls and 3 white balls. A single card is randomly selected from a standard deck. If the card is less than 5 (aces count as 1), a ball is drawn out of the first urn; otherwise a ball is drawn out of the second urn. If the drawn ball is red, what is the probability that it came out of the second urn?
(9/13)(2/5)
(9/13)(2/5)+(4/13)(4/5)
Given : One box has 7 red balls and 3 white balls;
a second box has 6 red balls and 4 white balls.
A pair of dice are tossed.
If the sum of the dice is less than five, a ball is selected from the first box, otherwise the ball is selected from the second box.
To Find : the probability of getting a red ball.
Solution:
Pair of Dice
Total possible outcomes = 6 * 6 = 36
Sum less than 5 are 2 , 3 and 4
(1 , 1) , ( 1 , 2) , ( 2 , 1) , ( 1 , 3) , (2 , 2) , ( 3 , 1)
Probability sum less than 5 = 6/36 = 1/6
Probability otherwise = 1 - 1/6 = 5/6
First box has 7 red balls and 3 white balls; P(R) = 7/10
Second box has 6 red balls and 4 white balls. P (R) = 6/10
probability of getting a red ball.
= (1/6) (7/10) + (5/6)(6/10)
= 7/60 + 30/60
= 37/60
Probability of getting a red ball is 37/60
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