Question

A die is thrown once ,find the probability of getting a number greater than 4 ​

Answer 1

Answer:

sample space of Die (s)=

(1,2,3,4,5,6)

Total no of outcomes are 6.

probablity of getting no greater than four(4)=2

P(E)=n(E)/n(s)

P(E)=2/6

P(E)=1/3

The probability of getting no greater than four is 1/3.

I hope it helps u..

Answer 2

Answer:

$\mathfrak{ \huge{ \green{ question : }}}$

ᴀ ᴅɪᴇ ɪs ᴛʜʀᴏᴡɴ ᴏɴᴄᴇ ,ғɪɴᴅ ᴛʜᴇ ᴘʀᴏʙᴀʙɪʟɪᴛʏ ᴏғ ɢᴇᴛᴛɪɴɢ ᴀ ɴᴜᴍʙᴇʀ ɢʀᴇᴀᴛᴇʀ ᴛʜᴀɴ 4 .

$\mathfrak{ \huge{ \green{ solution : }}}$

$\boxed{Probability \: of \: event \: to \: happen P(E) = \frac{Number \: of \: favourable \: outcomes}{Total \: Number \: of \: outcomes} }$

$Sample \: Space = {1, 2, 3, 4, 5, 6} \\ </p><p></p><p>Number \: of \: favourable \: event =2\\ </p><p></p><p>i.e. \: {5,6} \\ </p><p></p><p>Total number of outcomes = 6 \\ </p><p></p><p>Thus, Probability, P = \frac{2}{6} = \frac{1}{3} </p><p></p><p>$

$\mathfrak{ \huge{ \underline{ \underline{ \purple {⛄❤HOPE \: IT \: HELPS ❤⛄}}}}}$