1. An unbiased cubic die marked with 1,2,2,3,3,3 is rolled three times. What is the probability of getting a total score of 4 or 6?

2. In a batch of 10 articles, 4 are defective. 6 articles are taken from the batch for inspection. If more than 2 articles in this batch are defective, the whole batch is rejected. What is the probability that the batch will be rejected?

3. A draw 2 cards at random form a pack of 52 cards. After returning them to the pack and shuffling it, B draws 2 cards at random. What is the probability that there is exactly one common card?

3. Let S be the sample space and let E be the req. event, den

n(S) = (⁵²C₂)²

For the no. of elements in E, we frst choose a card (dat v want common) nd den from de remaining cards(i.e. 51) we choose 2 cards nd distribute dem among A and B in 2! ways.

So, n(E) - ⁵²C₁ . ⁵¹C₂ . 2!

Hence, P(E) = 50 / 663 (Ans)

1) (i) For obtaining 4, we need two 1s and one 2...............

         Hence, probability of obtaining a score of 4 = 3*1/6*1/6*1/3 = 1/36

     (ii)Six can be obtained in the following ways:-

a) one 1, one 2 and one 3

b) three 2s

      Hence probability of a score of 6 = 3*1/6*1/3*1/2 + 3*1/3*1/3 = 1/12 + 1/3 = 5/12

HENCE, PROBABILITY OF A SCORE OF 6 OR 4 = PROBABILITY OF OBTAINING 4 + PROBABILITY OF OBTAINING 6 = 5/12 + 1/36 = 16/36 =  4/9

2) Total number of ways of selecting the 6 articles = ¹⁰C₆ = a

     No of ways of selecting the six articles such that atmost two are defective =   ⁶C₆ + ⁴C₁ * ⁶C₅ + ⁴C₂ * ⁶C₄ = b

     Hence, probability of not being rejected = b/a

     And probability of being rejected = 1 - b/a

Batao Bhai......................Answer kya hai?................mera correct hai ya nahin?

2.  Six articles can be chosen from 10 articles in ¹⁰C₆ ways.

So, sample space = ¹⁰C₆

For the batch to be rejected more than 2 articles should be defective.

So, there are 2 cases ( since a max. of 4 atrciles are defective) -

(i) When 3 are defective out of 4 and 3 are non-defective out of 6.

(ii) When 4 are defective out of 4 and 2 are non-defective out of 6.

So, Req. Prob. = (⁶C₃ . ⁴C₃ / ¹⁰C₆) + (⁶C₂ . ⁴C₄ / ¹⁰C₆) = 19 / 42 (Ans)...............

thnkx....guys - deepak & transmigrator.

answers are

(1) 25/108

(2) 19/42

(3) 50/663

transmigrator u sol. to 1st question s X somewhere. plz check & post the correct one.

thnkx in dvnce.

(1) Probability of getting 1 = P(1) = 1/6

Probability of getting 2 = P(2) = 2/6

and, Probability of getting 3 = P(3) = 3/6

(i) For the sum to be 4, it can occur as (1,1,2)

Further these three can arrange in (3!/2!) ways (as 1 comes twice).

So, P(sum 4) = (1/6) (1/6) (2/6) (3!/2!) = 1 / 36

(ii) For the sum to be 6, it can occur in two ways as (2,2,2) and (1,2,3).

So, P(sum 6) = (2/6)³ + (1/6) (2/6) (3/6) (3!) = (1/27) + (1/6) = 11/54

So, req. Prob. = P(sum 4) + P(sum 6) = (1/36) + (11/54) = 25/108 (Ans).............