Question

kindly answer soon.....​

Answer 1

$\huge\red{\boxed{\sf AnSwER}}$

Total no. of cards in a pack =52

After removing red-colored - jack, queen, king, and aces,

No. of cards =52−8=44 and

No. of red cards =26−8=18.

$\huge\red{\boxed{\sf SoLuTiOn (1)}}$

No. of black queens =2

Therefore, E¹

( Selecting 1 out of 2 items) times out of 44E²

( Selecting 1 out of 44 items) a black queen is picked.

Let E be the event of getting a black queen from pack

We know that,

$Probability P(E)  \: = \: \: \: \frac{(No.of favorable outcomes) \: \: </h2><h2>}{total \: no \: of \: outcomes}$

$\huge\blue{\boxed{\sf \frac{1}{22}}}$

$\huge\red{\boxed{\sf Solution (2)}}$

No. of red cards =18

Therefore,

18E¹

( Selecting 1 out of 18 items) times out of

44E²

( Selecting 1 out of 44 items) a red card is picked.

Let E be the event of getting a red card from the pack

$Probability P(E)  \: = \: \: \: \frac{(No.of favorable outcomes) \: \: </h2><h2>}{total \: no \: of \: outcomes}$

$\huge\blue{\boxed{\sf \frac{9}{22}}}$

$\huge\red{\boxed{\sf SoLuTiOn (3)}}$

No. of black jack cards =2

Therefore,

2E¹

( Selecting 1 out of 2 items) times out of

44E²

( Selecting 1 out of 44 items) a black jack card is picked.

Let E be the event of getting a black jack card from the pack

We know that,

$Probability P(E)  \: = \: \: \: \frac{(No.of favorable outcomes) \: \: </h2><h2>}{total \: no \: of \: outcomes} </h2><h2>$

$\huge\blue{\boxed{\sf \frac{1}{22} }}$

$\huge\red{\boxed{\sf SoLuTiOn (4)}}$

No. of picture cards =12−6=6 ..... (As red cards are removed and there are 6 red picture cards )

Therefore,

6E¹

( Selecting 1 out of 6 items) times out of

44E²

( Selecting 1 out of 44 items) a picture card is picked.

Let E be the event of getting a picture card from pack

We know that,

$Probability P(E)  \: = \: \: \: \frac{(No.of favorable outcomes) \: \: </h2><h2>}{total \: no \: of \: outcomes}$

$= \frac{6}{44}$

$\huge\blue{\boxed{\sf \frac{3}{22} }}$