IIT JEE , AIEEE Forum - Mathematics , Algebra

dilip

PLz help me solve these probabiliy problems.I tried but my answers weren't matching with tht given in the book.below are answers from the book

:1.)A card is drawn from a pack of 52 cards.Find the probability that the card drawn is red or bears a number between 5 n 10 both exclusive.Ans..19/26

2)two cards are drawn at random from 52 playing cards.Find the probability tht one is a spade and the other is an aceAns..1/34

dilip

plz reply .rates r assured

sboosy

are u sure the answer is 19/26 or is it 17/26

iitkgp_bipin

Total no. of red cards among 52 cards = 26

There two different types of black colored cards each containing 6 cards numbered 5 to 10. no. of such cards = 12

no. of cards which are either red or are numbered between 5 to 10 = 26+12 = 38

total cards = 52

so required probability = 38/52 = 19/26

edited : i agree with sboosy, if 5 and 10 are not included required probability is 17/26.

punnima

ans for 2)

well prob=no.of fav. outcomes

ttl no. of outcomes

ttl no. of outcomes=⁵²C₂

no.of fav outcomes = (¹³C₁x⁴C₁) - (13)

the sub of 13 is because if d 1st one is an ace in spade the other one can be an ace only 4m diamond,heart and clover

th 13 =¹³C₁ x ¹C₁

sboosy

A-probability of getting red
B- probability of getting 6 or 7 or 8 or 9
P(A union B)=P(A)+P(B)-P(A intersect B)
P(A)=1/2
P(B)=16/52=4/13
Intersecct=P(A/B)*P(B)
1/2*16/52=2/13
so answer = 1/2+4/13-2/13
=17/26

sboosy

BIpin sir 5 and 10 not included

punnima

contd ans 2)....

ther4 reqd prob = [(13x4) / 1326 ]-[13/1326]
=39/1326
= 1/34
= 0.0294

punnima

1)

Ttl no of red cards amng 52 cards = 26

There r 2 diff. types of black colored cards each containing 4 cards numbered btw 5 to 10.

no. of such cards = 8

no. of cards which are either red or are numbered between 5 to 10 = 26+ 8 = 34

ttl cards = 52

so reqd prob. = 34/52 = 17/26

dilip

my answer too comes 17/26.this means tht some mistake has been done in the book.thanks for help.

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