The probability of getting 2 heads and 1 tail when three coins are tossed is 3 in 8. Find the odds of not getting 2 heads and 1 tail.

ANSWER: 5:8?

Three Coins are tossed. Find the probability that exactly 2 coins show heads if the first coin shows heads.?

ANSWERS: Could it be 1/4?

The first answer is correct.

For the second, once you have the first head there are four possibilities for the other two coins:

HH
TT
TH
HT

They must show one head for there to be two heads. What proportion do you find?

I hope this helps. Thanks for asking.

5:8

To find the odds of not getting 2 heads and 1 tail when three coins are tossed, we can subtract the probability of getting 2 heads and 1 tail from 1 (since the sum of the probabilities of all possible outcomes must equal 1).

Given that the probability of getting 2 heads and 1 tail is 3/8, we can calculate the odds of not getting 2 heads and 1 tail as:

Odds = (1 - Probability) / Probability

Odds = (1 - 3/8) / (3/8)

Odds = (8/8 - 3/8) / (3/8)

Odds = 5/8

Therefore, the odds of not getting 2 heads and 1 tail when three coins are tossed is 5:8.

Regarding the second question, if the first coin is known to show heads, the remaining two coins can each show either heads or tails. Since each coin has two possible outcomes, the total number of possible outcomes for the second and third coins is 2^2 = 4. Out of these 4 outcomes, only 1 outcome would result in exactly 2 coins showing heads.

Therefore, the probability of exactly 2 coins showing heads, given that the first coin shows heads, is 1 out of 4, which can be written as 1/4.

To find the odds of not getting 2 heads and 1 tail, we need to subtract the probability of getting 2 heads and 1 tail from 1.

Given that the probability of getting 2 heads and 1 tail is 3 in 8, the probability of not getting 2 heads and 1 tail can be found by subtracting 3/8 from 1.

1 - 3/8 = 8/8 - 3/8 = 5/8

Therefore, the odds of not getting 2 heads and 1 tail are 5:8.

For the second question, we are asked to find the probability of getting exactly 2 heads when the first coin shows heads. To solve this, let's break down the problem.

We know that the first coin shows heads. This means we have two more coins to consider, with two possible outcomes for each coin: heads or tails.

The remaining two coins can give us four possible outcomes: HH, HT, TH, and TT.

Out of these four outcomes, we are interested in the one that shows exactly 2 heads, which is HH.

Therefore, the probability of getting exactly 2 heads when the first coin shows heads is 1 out of 4, or 1/4.

For the second question, your answer of 1/4 is correct.