Wiley Edge Probability Quiz-1

Solution: The probability of drawing the first red ball is 6/10, and for the second red ball, it is 5/9. Therefore, the combined probability is  

\frac{6}{10} * \frac{5}{9} =  \frac{30}{90} = \frac{1}{3} .

Solution: The number of favorable outcomes corresponds to the number of ways we can arrange two heads (H) and three tails (T) in a sequence of five coin flips. This can be calculated using the binomial coefficient formula, also known as "n choose k".

The binomial coefficient formula is given by:

\frac{n!}{(k!(n-k)!)}

where n is the total number of trials (in this case, the number of coin flips) and k is the number of successful outcomes (in this case, the number of heads).

Using this formula, we can calculate the number of favorable outcomes:

Favorable outcomes = \frac{5!}{(2!(5-2)!)} = \frac{5!}{(2!3!)} = \frac{(5 * 4) }{(2 * 1)} = 10

The total number of possible outcomes is given by 2 to the power of the total number of trials, since each coin flip has two possible outcomes (head or tail):

Total possible outcomes = 2^{5} = 32

Therefore, the probability of getting exactly two heads and three tails is:

Probability = (Number of favorable outcomes) / (Total number of possible outcomes) =  \frac{10}{32} = \frac{5}{16}

So, the probability is 5/16.

 

Solution: The probability of drawing two red balls is 

\frac{7}{15} *  \frac{6}{14}   = \frac{42}{210} =  \frac{21}{105} . Similarly, the probability of drawing two white balls or two green balls is 10/105 and 3/105, respectively. 

Adding these probabilities together, we get 

\frac{21}{105} +  \frac{10}{105} +  \frac{3}{105} =  \frac{34}{105} =  \frac{2}{7}.

Solution: The probability of drawing a red ball first and a blue ball second is (4/9) * (3/8) = 12/72 = 1/6. Similarly, the probability of drawing a blue ball first and a red ball second is (3/9) * (4/8) = 12/72 = 1/6. Adding these probabilities together, we get 1/6 + 1/6 = 2/6 = 1/3. 

Solution: The odd prime numbers on a six-sided die are 3 and 5. There are three rolls, so the probability of not getting an odd prime number on a single roll is:

P(not odd prime) = 1 - P(odd prime)

= 1 - (2/6)

= 1 - 1/3

= 2/3

Since the rolls are independent events, we can multiply the probabilities for each roll:

P(not odd prime in three rolls) = (2/3) * (2/3) * (2/3)

= 8/27

Finally, we subtract this probability from 1 to find the probability of getting an odd prime number at least once:

P(odd prime at least once) = 1 - P(not odd prime in three rolls)

= 1 - 8/27

= 19/27

Therefore, the probability of getting an odd prime number at least once in three rolls of a fair six-sided die is 19/27.

 

Solution: The probability of drawing two red balls and one white ball is (7/16) * (6/15) * (5/14) = 210/3360 = 35/560 = 7/112. 

Solution: The number of ways to select one queen from 4 queens is 4, and the number of ways to select one king from 4 kings is also 4.

Therefore, the number of favorable outcomes is 4 * 4 = 16.

The total number of cards in a standard deck is 52.

Thus, the total number of possible outcomes is 52.

Therefore, the probability of drawing a queen and a king together is:

P(queen and king) = favorable outcomes / total outcomes

= 16 / 52

= 4 / 13

Therefore, the probability of drawing a queen and a king together from a deck of playing cards is 4/13.

Solution: The probability of drawing three black balls is (10/24) * (9/23) * (8/22) = 720/12144 = 5/84. The probability of drawing two black balls and one ball of a different color can be calculated in three ways: (10/24) * (9/23) * (16/22), (10/24) * (8/23) * (16/22), and (9/24) * (10/23) * (16/22). Adding these probabilities together, we get (5/84) + (5/84) + (5/84) = 15/84 = 5/28. Therefore, the probability of drawing at least two black balls is (5/84) + (5/28) = 65/336.

Solution: The probability of drawing three red balls is (6/13) * (6/13) * (6/13) = 216/2197. Similarly, the probability of drawing three blue balls or three green balls is (4/13) * (4/13) * (4/13) = 64/2197. Adding these probabilities together, we get (216/2197) + (64/2197) + (64/2197) = 344/2197 = 4/27. 

 

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.