Question

Probability
Class 10
•Please answer both the questions.​

Answer 1

Your Answer:

The possibility of dice are shown in the figure 1

5)

i) A multiple of 3 as the sum

So, The possibilities are

E = {(1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6,3), (6,6)}

So, Favourable Outcome = 12

And Total Outcome = 36

P(E) = Favourable Outcome / Total Outcome = 12/36 = 1/3

ii) Doublet

So, The possibilities are

E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

So, Favourable Outcome = 6

And Total Outcome = 36

P(E) = Favourable Outcome / Total Outcome = 6/36 = 1/6

iii) A total of at least of 10

So, The possibilities are

E = {(4,6),  (5,5), (5,6), (6,4), (6,5), (6,6)}  

So, Favourable Outcome = 6

And Total Outcome = 36

P(E) = Favourable Outcome / Total Outcome = 6/36 = 1/6

6)

Total Card remaining = 52 - 3 = 49

i) Diamond cards left = 10 - 3 = 10

Total Card = 49

P(E) = 10/49

ii) One Jack is removed  

So, Jack Remaining = 4 - 1 = 3

Total cards = 49

P(E) = 3/49

iii) There is One 10 of hearts

Total Card = 49

P(E) = 1/49

iv) Face card left = 12 - 3 = 9

Total Cards = 49

P(E) = 9/49

Answer 2

Answer

$\sf Total\:number\:of\:outcomes\:when\:two\:dice\:thrown=6^2=36$

$\sf Possible\:outcomes\:when\:two\:dice\:are\:rolled:$

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

$\sf 1.(i)\: We\:know\:that,\sf From\:the\:above\:observation,there\:are\:12\:outcomes\:which\:generate\\\:a\:multiple\:of\:3\:as\:the\:sum.$

$\sf 12\:outcomes\:which\:generate\:multiple\:of\:3\:as\:the\:sum\:are:$

$\sf (1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6).$

$\sf \therefore \boxed{\sf Probability=\dfrac{12}{36}=\dfrac{1}{3}}$

$\sf (ii) Doublet\:means\:when\:both\:dice\:matches\:with\:each\:other.$

$\sf So,favourable\:outcomes\:are:$

$\sf (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) =6\:favorable\:outcomes.$

$\sf \therefore \boxed{\sf Probability =\dfrac{6}{36}=\dfrac{1}{6}}$

$\sf(iii) Favorable\:outcomes\:are:$

$\sf (4,6), (5,5), (5,6), (6,4), (6,5), (6,6)=6\:favorable\:outcomes.$

$\sf \therefore \boxed{\sf Probability =\dfrac{6}{36}=\dfrac{1}{6}}$

$\sf 2.King\:queen\:and\:jack\:of\:diamonds\:are\:removed\:from\:a\:52\:pack\:of\:cards.$

$\sf Remaining\:cards=52-3=49\:cards.$

$\sf (i) We\:know\:that,there\:are\:13\:cards\:of\:diamonds.When\:3\:cards\:are\:removed\:then,$

$\sf Remaining\:cards=13-3=10\:cards$

$\sf \therefore \boxed{\sf Probability=\dfrac{10}{49}}$

$\sf (ii) We\:know\:that\:there\:are\:4\:jacks\:in\:cards.\:When\:1\:jack\:is\:removed\:then,$

$\sf Remaining\:jacks=4-1=3\:jacks.$

$\sf \therefore \boxed{\sf Probability=\dfrac{3}{49}}$

$\sf (iii)We\:know\:that\:there\:are\:13\:cards\:of\:hearts.\:So,$

$\sf \therefore \boxed{\sf Probability=\dfrac{1}{49}}$

$\sf(iv) We\:know\:that\:there\:are\:12\:face\:cards.But\:3\:cards\:which\:are\:face\:cards\:are\:removed.$

$\sf Remaining\:face\:cards=12-3=9\:face\:cards.$

$\sf \therefore \boxed{\sf Probability=\dfrac{9}{49}}$