Wiley Edge Probability Quiz-2

Solution: The probability of drawing a red ball first is 8/15. Given that the first ball drawn was red, the probability of drawing three blue balls is (7/14) * (6/13) * (5/12) = 35/156 = 7/31. 

 

Solution: The probability of drawing a red ball and a non-white ball is given by (7/16) * (11/15) + (5/16) * (9/15) + (4/16) * (12/15) = 77/240 + 45/240 + 48/240 = 170/240 = 17/24. 

 

Solution: The probability of getting an even number on a fair six-sided die is 3/6 = 1/2, and the probability of getting an odd number is also 1/2. Using the binomial coefficient, the probability of getting exactly two even numbers and four odd numbers is C(6, 2) * (1/2)^2 * (1/2)^4 = 15/64.

Solution: The probability of drawing two red balls and one non-red ball is given by 

(5/12) * (4/11) * (8/10) = 160/1320 = 4/33. 

Similarly, the probability of drawing two blue balls and one non-blue ball is (4/12) * (3/11) * (9/10) = 108/1320 = 9/110. Adding these probabilities together, we get (4/33) + (9/110) = 19/132 + 9/110 = 209/1320 = 19/120. 

Solution: To calculate the probability of rolling a 6 twice on a fair six-sided die, we need to consider the probability of rolling a 6 on a single roll and multiply it by itself for two consecutive rolls.

The probability of rolling a 6 on a single roll is \frac{1}{6} , since there is only one favorable outcome (rolling a 6) out of six possible outcomes (rolling numbers 1 through 6).

To find the probability of rolling a 6 twice in a row, we multiply these probabilities together:

P(rolling a 6 twice) = P(rolling a 6 on the first roll) * P(rolling a 6 on the second roll)

= \frac{1}{6} * \frac{1}{6}  

= \frac{1}{36}

Therefore, the probability of rolling a 6 twice on a fair six-sided die is 1/36.

 

Solution: To find the probability of drawing two cards that are either both aces or one ace and one jack, we need to calculate the probability of each case and add them together.

Case 1: Both cards are aces

There are four aces in a standard deck of 52 cards. After the first ace is drawn, there are three aces remaining in the deck out of 51 cards. Thus, the probability of drawing two aces is:

P(both aces) = \frac{4}{52} * \frac{3}{51} =  \frac{1}{221}

Case 2: One ace and one jack

There are four aces and four jacks in a deck of 52 cards. After the first card is drawn, there are three aces and four jacks remaining out of 51 cards. The probability of drawing one ace and one jack can be calculated as:

P(ace and jack) = (\frac{4}{52} * \frac{4}{51}) + (\frac{4}{52} * \frac{4}{51})= \frac{8}{663} 

Finally, we add the probabilities of both cases:

P(both aces or ace and jack) = P(both aces) + P(ace and jack) = \frac{1}{221} + \frac{8}{663}   = \frac{11}{663}

Therefore, the probability of drawing two cards that are either both aces or one ace and one jack is 11/663

Solution: The number of ways to select one green ball from 5 green balls is 5, and the number of ways to select one blue ball from 5 blue balls is also 5.

Therefore, the number of favorable outcomes is 5 * 5 = 25.

The total number of balls in the bag is 5 + 5 + 5 = 15.

Thus, the total number of possible outcomes is 15.

Therefore, the probability of drawing a green and blue ball together is:

P(green and blue) = favorable outcomes / total outcomes

=  \frac{25}{15}  

=  \frac{5}{3}

≈ 0.833

Therefore, the probability of drawing a green and blue ball together from the given bag is approximately 0.833 or 5/3.

 

Solution: When two fair coins are tossed simultaneously, there are four possible outcomes: HH, HT, TH, and TT, where H represents a head and T represents a tail.

Out of these four outcomes, only two of them have a tail in one coin and a head in the other: HT and TH.

Therefore, the probability of getting a tail in one coin and a head in the other at the same time is 2 out of 4, or 1/2.

 

Solution: There are 9 blue or green balls out of a total of 15 balls. Thus, the probability is 9/15 or 3/5. 

Solution: A standard deck of 52 playing cards consists of 26 black cards (13 spades and 13 clubs), 26 red cards (13 hearts and 13 diamonds), and 2 joker cards.

To calculate the probability of drawing a black card or a joker, we need to determine the number of favorable outcomes and the total number of possible outcomes.

Favorable outcomes: The number of black cards plus the number of joker cards is 26 + 2 = 28.

Total possible outcomes: The total number of cards in a standard deck is 52.

Therefore, the probability of drawing a black card or a joker is 28/52, which simplifies to 7/13.