1. On a single toss of a fain coin, the probability of heads is 0.5 and the prob

Question

1. On a single toss of a fain coin, the probability of heads is 0.5 and the probability of tails is 0.5. If you toss a coin twice and get heads on the first toss, are you guaranteed to get tails on the second toss? Explain.

2. You are given the information that P(A) = 0.30 and P(A) = 0.40. Do you have enough information to compute P(A or B) and/or to compute P(A and B)? Explain. If you know that events A and B are mutually exclusive, do you have enough information to compute P(A or B)? Explain. If you know that events A and B are independent, do you have enough information to compute P(A and B)? Explain.

3. You need to know the number of different arrangements possible for five distinct letters. You decide to use the permutations rule, but your friend tells you to use 5!. Who is correct? Explain.

Explanation / Answer

1) On a single toss of a fain coin, the probability of heads is 0.5 and the probability of tails is 0.5.

If we toss a coin twice and get heads on the first toss, are you guaranteed to get tails on the second toss?

solution: No, the probability of heads on the second toss is 0.50 regardless of the outcome on the first toss.

Suppose, we toss two coins, there can be two heads, two tails, or a head and a tail. It is tempting to say that there are three equally possible outcomes.

Lets see explanation:

The probability that the coin when tossed turns up heads is 1/2. This means that when the coin is tossed, the probability of tails is 1 – 1/2 = 1/2.

When the coin is tossed twice, the probability of getting only tails is 1/2* 1/2 = 1/4. The probability of getting at least one heads is 1 – 1/4 = 3/4.

Now, we have to remember that the probability of getting a heads equal to 1/2 does not mean that for every two tosses, one is definitely going to be heads and the other tails. Though if we continue to toss the coin a very large number of times, we will find that the number of times it turns up tails is very close to the number of times it turns up heads.

2)

i) We are given the information that P(A) = 0.30 and P(B) = 0.40.

P(A or B) = P(A) + P(B)-P(A and B).
Since P(A) = 0.3, P(B) = 0.4, but P(Aand B) is not given, you do not have enough information to find P(AorB).

Similarly,

P(A and B) = P(A) + P(B)-P(A or B).
Since P(A) = 0.3, P(B) = 0.4, but P(A or B) is not given, you do not have enough information to find P(A and B).

ii)  If you know that A and B are mutually exclusive, then
P(A or B) = P(A) +P(B)
Since P(A) = 0.3 and P(B) = 0.4, you have enough information to find P(AorB)

if you know that A and B are mutually exclusive.

iii) If we know A and B are independents then we can calculate P(A and B)

3)

There are 5 choices for the first letter. Having chosen a first letter there are 4 choices for the second letter. Therefore there are 5 * 4 = 20 choices for the first two letters. Continuing this reasoning gives the following number of arrangements for the five letters taken five at a time:

5×4×3×2×1=5!

If we use the formula for permutations we get

5P5 = 5!

For example:How many different letter arrangements can be made from the letters: FLUKE

all the letters are different so we can make 5! = 120 arrangements .

You can go by using Permutation rule it is correct.

Your freind answer is also correct.

But, always go step by step.

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