This is a multi-part question in the textbook answered separately on Clay6.com

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

related to an answer for:
Find the probability distribution of number of heads in two tosses of a coin.

0 votes

- Once we write down the sample space for the problem we can find the number of heads or tails for the favorable outcomes in each case and find their probabilities and arrange as a probability distribution.

Given: three simultaneous tosses of three coins, sample space can be represented as follows; S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

All these events are equally likely and have the same probability = $\large\frac{1}{8}$

Let X be the number of tails in the tosses, such that $\rightarrow$ X(HHH) = 0, X(HHT) = 1, X(HTH) = 1, X(HTT) = 2, X(THH) = 1, X(THT) = 2, X(TTH) = 2 and X(TTT) = 3.

Therefore, X can assume a value of 0, 1, 2 and 3. It follows that:

P (X = 0) = P (HHH) = $\large\frac{1}{8}$

P (X = 1) = P (HHT, HTH, THH) = $\large \frac{3}{8}$

P (X = 2) = P (HTT, THT, TTH,) = $\large\frac{3}{8}$

P (X = 3) = P (TTT) = $\large\frac{1}{8}$

Therefore we can write the probability distribution as follows:

$\begin{matrix} \textbf{X} & 0 &1 &2&3 \\ \textbf{P(X)} &\large \frac{1}{8} &\large \frac{3}{8} &\large \frac{3}{8} &\large\frac{1}{8}\end{matrix}$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...