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 Question from debbie, a student: What is the probability of getting 6 out of 10 (a,b,c,d,e) questions correct, by random selection?

Hi Debbie. I'll show you with a similar test.

What is the probability of getting 2 out of 4 (a, b, c, d) questions correct by random selection?

It could be any set of two out of the four questions, so we start with 4C2 (that's questionsCcorrect) possible choices.

For each of those, we know that each of the questions we got wrong had three possible choices of "wrong" answer. So there are 3(4-2) (that's [wrong choices](wrong questions) ) choices for wrong answers.

That means that there are 4C2 × 32 ways of getting 2 out of 4 on this test.

Now how many ways are there of answering the test in all? Since each question has 4 choices and there are 4 questions, there are 44 (that's choicesquestions) possible ways of answering the test.

Thus, the probability is (the number of ways we can get 2 out of 4) divided by the (number of possible ways of answering in total).

So the probability for my situation is ( 4C2 × 32 ) / 44 = (6 × 9) / 256 = 54 / 256 = 27 / 128. You can answer your problem the same way.

Here's the proof that my answer is correct: I will write out all the possibilities.

 BAAA : 0 BAAB : 0 BAAC : 0 BAAD : 1 BABA : 0 BABB : 0 BABC : 0 BABD : 1 BACA : 1 BACB : 1 BACC : 1 BACD : 2 BADA : 0 BADB : 0 BADC : 0 BADD : 1 BBAA : 1 BBAB : 1 BBAC : 1 BBAD : 2 BBBA : 1 BBBB : 1 BBBC : 1 BBBD : 2 BBCA : 2 BBCB : 2 BBCC : 2 BBCD : 3 BBDA : 1 BBDB : 1 BBDC : 1 BBDD : 2 BCAA : 0 BCAB : 0 BCAC : 0 BCAD : 1 BCBA : 0 BCBB : 0 BCBC : 0 BCBD : 1 BCCA : 1 BCCB : 1 BCCC : 1 BCCD : 2 BCDA : 0 BCDB : 0 BCDC : 0 BCDD : 1 BDAA : 0 BDAB : 0 BDAC : 0 BDAD : 1 BDBA : 0 BDBB : 0 BDBC : 0 BDBD : 1 BDCA : 1 BDCB : 1 BDCC : 1 BDCD : 2 BDDA : 0 BDDB : 0 BDDC : 0 BDDD : 1
 CAAA : 0 CAAB : 0 CAAC : 0 CAAD : 1 CABA : 0 CABB : 0 CABC : 0 CABD : 1 CACA : 1 CACB : 1 CACC : 1 CACD : 2 CADA : 0 CADB : 0 CADC : 0 CADD : 1 CBAA : 1 CBAB : 1 CBAC : 1 CBAD : 2 CBBA : 1 CBBB : 1 CBBC : 1 CBBD : 2 CBCA : 2 CBCB : 2 CBCC : 2 CBCD : 3 CBDA : 1 CBDB : 1 CBDC : 1 CBDD : 2 CCAA : 0 CCAB : 0 CCAC : 0 CCAD : 1 CCBA : 0 CCBB : 0 CCBC : 0 CCBD : 1 CCCA : 1 CCCB : 1 CCCC : 1 CCCD : 2 CCDA : 0 CCDB : 0 CCDC : 0 CCDD : 1 CDAA : 0 CDAB : 0 CDAC : 0 CDAD : 1 CDBA : 0 CDBB : 0 CDBC : 0 CDBD : 1 CDCA : 1 CDCB : 1 CDCC : 1 CDCD : 2 CDDA : 0 CDDB : 0 CDDC : 0 CDDD : 1
 DAAA : 0 DAAB : 0 DAAC : 0 DAAD : 1 DABA : 0 DABB : 0 DABC : 0 DABD : 1 DACA : 1 DACB : 1 DACC : 1 DACD : 2 DADA : 0 DADB : 0 DADC : 0 DADD : 1 DBAA : 1 DBAB : 1 DBAC : 1 DBAD : 2 DBBA : 1 DBBB : 1 DBBC : 1 DBBD : 2 DBCA : 2 DBCB : 2 DBCC : 2 DBCD : 3 DBDA : 1 DBDB : 1 DBDC : 1 DBDD : 2 DCAA : 0 DCAB : 0 DCAC : 0 DCAD : 1 DCBA : 0 DCBB : 0 DCBC : 0 DCBD : 1 DCCA : 1 DCCB : 1 DCCC : 1 DCCD : 2 DCDA : 0 DCDB : 0 DCDC : 0 DCDD : 1 DDAA : 0 DDAB : 0 DDAC : 0 DDAD : 1 DDBA : 0 DDBB : 0 DDBC : 0 DDBD : 1 DDCA : 1 DDCB : 1 DDCC : 1 DDCD : 2 DDDA : 0 DDDB : 0 DDDC : 0 DDDD : 1

Now you try it with your problem.

Cheers,
Stephen La Rocque

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