Three unbiased coins are tossed together. The probability of getting a...
Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.
P(E) = n(E) / n(S)
= 4/8= 1/2
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Three unbiased coins are tossed together. The probability of getting a...
HTH HHT THH HHH
So we have
4/8=1/2
hence A is right option
Three unbiased coins are tossed together. The probability of getting a...
To solve this problem, we can use the concept of probability. Let's break it down step by step:
Step 1: Identify the possible outcomes
When three unbiased coins are tossed together, there are a total of 2^3 = 8 possible outcomes. Each coin can either land heads or tails, and there are three coins in total.
The possible outcomes are:
- HHH (3 heads)
- HHT (2 heads, 1 tail)
- HTH (2 heads, 1 tail)
- THH (2 heads, 1 tail)
- HTT (1 head, 2 tails)
- THT (1 head, 2 tails)
- TTH (1 head, 2 tails)
- TTT (3 tails)
Step 2: Determine the favorable outcomes
To find the probability of getting at most two heads, we need to determine the number of outcomes where we get 0, 1, or 2 heads.
The favorable outcomes are:
- TTT (0 head)
- HTT (1 head)
- THT (1 head)
- TTH (1 head)
- HHT (2 heads)
- HTH (2 heads)
- THH (2 heads)
There are a total of 7 favorable outcomes.
Step 3: Calculate the probability
The probability is defined as the number of favorable outcomes divided by the total number of possible outcomes.
Probability of getting at most two heads = favorable outcomes / total outcomes
= 7 / 8
Therefore, the correct answer is option 'A' - 7/8.
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