Question

let x denote the number of heads in a single toss of 4 fair coins determine p(x 2)​

Answer 1

Given : x denote the number of heads in a single toss of 4 fair coins

To Find :  P(x)

Solution:

a single toss of 4 fair coins

=> 2 * 2 * 2 * 2  = 16  out comes

( HHHH , HHHT , HHTH  , HHTT , HTHH , HTHT , HTTH , HTTT ,

THHH , THHT , THTH  , THTT , TTHH , TTHT , TTTH , TTTT , )

                                 f            cases

P (0) = 0 heads  =      1          TTTT

P(1) = 1 Head              4        HTTT ,  THTT TTHT , TTTH

P(2) = 2 Head             6        HHTT  , HTHT ,  HTTH ,THHT , THTH  , TTHH ,

P(3) = 3 Head             4         HHHT , HHTH  , HTHH , THHH ,

P(4) = 4 Head             1         HHHH

P(0)  = 1/16  

P(1) = 4/16 = 1/4  

P(2)  = 6/16 = 3/8 ,

P(3) = 4/16 = 1/4

P(4)  = 1/16

P (2)  = 6/16  =  3/8  

Learn More:

A r.v. X has the given probability distribution, Find the value of k and ...

brainly.in/question/6536531

Assume that adults have iq scores that are normally distributed with ...

brainly.in/question/11133397

1) If the probability of a machine producing a defective part is 0.05 ...

brainly.in/question/18065898

Answer 2

$\underline{\textbf{Given:}}$

$\textsf{X denotes the number of heads in a single die of 4 fair coins}$

$\underline{\textbf{To find:}}$

$\textsf{P(x=2)}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Concept used:}}$

$\textsf{The probability mass function of binomial distribution is}$

$\boxed{\mathsf{P(X=x)=n_{C_x}\;p^x\;q^{n-x}}}$

$\mathsf{x=0,1,2,\;.\;.\;.n}$

$\textsf{where, n is number of trials}$

$\textsf{p-probability of success}$

$\textsf{q-probability of failure}$

$\mathsf{Here,\;n=4,}$

$\mathsf{p=P(getting\;head)=\dfrac{1}{2}}$

$\mathsf{q=P(getting\;tail)=\dfrac{1}{2}}$

$\mathsf{The\;probability\;function\;is}$

$\mathsf{P(X=x)=4_{C_x}\;\left(\dfrac{1}{2}\right)^x\;\left(\dfrac{1}{2}\right)^{4-x}}$

$\mathsf{P(X=x)=4_{C_x}\;\left(\dfrac{1}{2}\right)^4}$

$\implies\mathsf{P(X=2)=4_{C_2}\;\left(\dfrac{1}{2}\right)^4}$

$\implies\mathsf{P(X=2)=\dfrac{4{\times}3}{1{\times}2}\left(\dfrac{1}{16}\right)}$

$\implies\mathsf{P(X=2)=2{\times}3\left(\dfrac{1}{16}\right)}$

$\implies\boxed{\mathsf{P(X=2)=\dfrac{3}{8}}}$

$\underline{\textbf{Find more:}}$