Determine the probability p, for each of the following events.

(a) An odd number appears in a single toss of a fair die.

(b) At least one head appears in two tosses of a fair coin.

(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.

(d) The sum of 6 appears in a single toss of a pair of fair dice.

4 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
P
8 months ago

This is a long answer type question as classified in NCERT Exemplar

(a)Possibleoutcomeofasinglethrowofdie S = { 1 , 2 , 3 , 4 , 5 , 6 } o u t o f w h i c h 1 , 3 , 5 a r e o d d R e q u i r e d p r o b a b i l i t y = 3 6 = 1 2 ( b ) W h e n a f a i r c o i n i s t o s s e d t w i c e , t h e n t h e s a m p l e s p a c e S = { H H , H T , T H , T T } P r o b a b i l i t y o f g e t t i n g a t l e a s t o n e h e a d ( H H , H T , T H ) = 3 4 ( c ) F a v o u r a b l e e v e n t s a r e 4 k i n g s + 2 o f h e a r t s + 3 o f s p a d e s = 4 + 1 + 1 = 6 = 6 5 2 = 3 2 6 ( d ) W h e n a p a i r o f d i c e i s r o l l e d , t h e n t o t a l n u m b e r o f s a m p l e s p a c e = 3 6 o u t o f w h i c h ( 1 , 5 ) , ( 5 , 1 ) , ( 2 , 4 ) , ( 4 , 2 ) a n d ( 3 , 3 ) a r e t h e f a v o u r a b l e e v e n t s R e q u i r e d p r o b a b i l i t y = 5 3 6 .

Thumbs Up IconUpvote Thumbs Down Icon

Similar Questions for you

  | 1 2 2 i + 1 | = α ( 1 2 2 i ) + β ( 1 + i )  

9 4 + 4 = α ( 1 2 2 i ) + β ( 1 + i )

5 2 = α ( 1 2 ) + β + i ( 2 α + β )             

α 2 + β = 5 2      ...(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

α 2 + 2 α = 5 2

5 2 α = 5 2            

a =

...Read more

Variance = x 2 n ( x ¯ ) 2  

6 0 2 + 6 0 2 + 4 4 2 + 5 8 2 + 6 8 2 + α 2 + β 2 + 5 6 2 8 = ( 5 8 ) 2 = 6 6 . 2            

7 2 0 0 + 1 9 3 6 + 3 3 6 4 + 4 6 2 4 + 3 1 3 6 + α 2 + β 2 8 = 3 3 6 4 = 6 6 . 2             

2 5 3 2 . 5 + α 2 + β 2 8 3 3 6 4 = 6 6 . 2            

α2 + β2 = 897.7 × 8

= 7181.6

...Read more

Start with

(1) E ¯ : 6 ! 2 ! = 3 6 0  

(2)    G E ¯ : 5 ! 2 ! , G N ¯ : 5 ! 2 !  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

...Read more

( 1 + x ) 1 1 = 1 1 C 0 + 1 1 C 1 x + 1 1 C 2 x 2 + . . . . . + 1 1 C 1 1 x 1 1

= 2 1 2 2 2 4 1 2
= 2 1 2 2 6 1 2 = 4 0 7 0 1 2 = 2 0 3 5 6 = m n
m + n = 2035 + 6 = 2041

 

f ( x ) = { 2 + 2 x , x ( 1 , 0 ) 1 x 3 , x [ 0 , 3 )

g ( x ) = { x , x [ 0 , 1 ) x , x ( 3 , 0 )   ->g(x) = |x|, x Î (–3, 1)

f ( g ( x ) ) = { 2 + 2 | x | , | x | ( 1 , 0 ) x ? 1 | x | 3 , | x | [ 0 , 3 ) x ( 3 , 1 )            

f ( g ( x ) ) = { 1 x 3 , x [ 0 , 1 ) 1 + x 3 , x ( 3 , 0 )

Range of fog(x) is [0, 1]

            

            Range of fog(x) is [0, 1]

...Read more

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 2025

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering