What is the probability that 2 or more numbers out of 4 are among the drawn lottery numbers (6 out of 49 without replacement)?

Question

Meadow Knox · Accepted Answer

1. Draw four numbers out of six. All four balls must belong to the six chosen balls, so the probability of this happening equals:      6    49    ⋅      5    48    ⋅      4    47    ⋅      3    46  2. Draw three numbers out of six. One way to do this, is by first choosing three balls belonging to the six chosen balls, followed by one ball which does not belong to these balls. This probability equals:      6    49    ⋅      5    48    ⋅      4    47    ⋅      43    46  Since there are four different turns in which we can choose a ball not belonging to the six chosen balls, the probability of selecting three balls out of six equals:                 (              4      1                  )        ⋅      6    49    ⋅      5    48    ⋅      4    47    ⋅      43    46  3. Draw two numbers out of six. This time there are                 (              4      2                  )        =  6 combinations of turns in which we can choose a ball not belonging to the six chosen balls, so this happens with probability:                (              4      2                  )        ⋅      6    49    ⋅      5    48    ⋅      43    47    ⋅      42    46  Adding this all up, the probability of choosing at least two correct numbers equals:            6      ⋅      5      ⋅      4      ⋅      3      +      4      ⋅      6      ⋅      5      ⋅      4      ⋅      43      +      6      ⋅      6      ⋅      5      ⋅      43      ⋅      42              49      ⋅      48      ⋅      47      ⋅      46        ≈  0.0681

Richardtb · Accepted Answer

There are 6 winning numbers and 43 &quot;non-winning&quot; numbers.To get only 2 winning numbers out of the 4 you choose, using the standard hypergeometric formula for choosing w/o replacement, there are                 (              6      2                  )                      (              43      2                  )       ways out of a total of                 (              49      4                  )       waysProceeding similarly,   P  (  2  ,  3    or    4    winning  )  =                                                    (                                6            2                                          )                                                          (                                43            2                                          )                          +                                        (                                6            3                                          )                                                          (                                43            2                                          )                          +                                        (                                6            4                                          )                                                          (                                43            0                                          )                                                        (                          49          4                                  )                      =            515      7567        ≈  0.06806

What is the probability that 2 or more numbers out

Answered question

Answer & Explanation

New Questions in Precalculus