IIT JEE , AIEEE Forum - Mathematics , Algebra

ayshwarya

2 players A&B throw alterenatively a pair of dice .A wins if he throws 8 before B throws &Bthrows 7 & wins before A throws 8 . if A starts the game, then d ratio P(A):P(B) is (ans 30:31 )

there are 4 defective item in a lot of 15 items.d items are selected one by one at random without replacement till d last defective item is drawn.d probability dat d 10th item examined is d llast defective item is........................(ans 4/65)

akhil_o

ex[plained below:plz see

kishore.subramanian.b

Two dice are thrown Hence n(S) = 36 n(8) =5 n(7) =6

Let probability of A wining be P(W₁) = 5/36 P(L₁) =31/36

Let probability of B wining be P(W₂) = 6/36 P(L₂) =30/36

If A wins in I throw = 5/36

II throw = 5/36 + 31/36 *30/36* 5/36

III throw = 5/36 + 31/36 *30/36* 5/36 + (31/36)² *(30/36^{)2 *} 5/36

and so on ...........

Hence P(A) = 5/36 + 31/36 *30/36* 5/36 + (31/36)² *(30/36^{)2 *} 5/36 +............

=5/36 * 1/(1-31/36 *30/36) = 30/61

If B wins in I throw = 31/36 *6/36

II throw = 31/36 *6/36 + (31/36)² *30/36* 6/36

III throw = 31/36 *6/36 + (31/36)² *30/36* 6/36+ (31/36)³ *(30/36)^{2 *} 6/36

and so on ...........

Hence P(B) = 31/36 *6/36 + (31/36)² *30/36* 6/36+ (31/36)³ *(30/36)^{2 *} 6/36+....

=31/36 *6/36 *1/(1-31/36 *30/36) = 31/61

Hence P(A) : P(B) = 30 : 31

Rate me if correct............

kishore.subramanian.b

please explane the 2 question in detail akhil_o .........

akhil_o

ok ...as u wish...
for the 10th piece lifted to be the 4th faulty piece..
we have to have picked 3 bad ones earlier...
so we choose any 3 out of 9 places in 9C3 ways
also the probability of this happening is
(4/15)(3/14)(2/13)(1/12) as initially there are 4 pieces faulty, then 3, then 2 then lastly 1
note that if we have such a combination for any no. of possibilities, the non-defective poieces reagrrange themselves and there is no change in the probability

hence required answer=9C3(4/15)(3/14)(2/13)(1/12)
=4/65

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