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Question
i am going to toss three coins simultaneously. what is the probability that exactly two of the coins will land heads? a. $\frac{1}{6}$ b. $\frac{1}{4}$ c. $\frac{1}{3}$ d. $\frac{3}{8}$
Step1: Calculate total outcomes
Each coin has 2 possible outcomes. For 3 coins, the total number of outcomes is $2\times2\times2 = 2^3=8$ by the multiplication - principle.
Step2: Calculate favorable outcomes
Use the combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n = 3$ (number of coin - tosses) and $k = 2$ (number of heads).
$C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=\frac{3\times2!}{2!×1}=3$.
Step3: Calculate probability
The probability $P$ of an event is given by $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. So $P=\frac{3}{8}$.
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None of the provided options are correct. The correct probability is $\frac{3}{8}$.