Question

Out of 7 lottery tickets 3 are prize-winning tickets. If someone buys 4 tickets what is the probability of winning at least two prizes?

Answer 1

`P=22/35`

Explanation:

So, we have `3` winning and `4` non-winning tickets among `7` tickets available.

Let's separate the problem into four independent mutually exclusive cases:

(a) there are `0` winning tickets among those `4` bought
(so, all `4` bought tickets are from a pool of `4` non-winning tickets)

(b) there is `1` winning ticket among those `4` bought
(so, `3` bought tickets are from a pool of `4` non-winning tickets and `1` ticket is from a pool of `3` winning tickets)

(c) there are `2` winning tickets among those `4` bought
(so, `2` bought tickets are from a pool of `4` non-winning tickets and `2` tickets are from a pool of `3` winning tickets)

(d) there are `3` winning tickets among those `4` bought
(so, `1` bought ticket is from a pool of `4` non-winning tickets and `3` tickets are from a pool of `3` winning tickets)

Each of the above events has its own probability of occurrence. We are interested in events (c) and (d), the sum of the probabilities of their occurrence is what the problem is about. These two independent events constitute the event "winning at least two prizes". Since they are independent, a combined event's probability is a sum of its two components.

Probability of event (c) can be calculated as a ratio of the number of combinations of `2` bought tickets are from a pool of `4` non-winning tickets and `2` tickets are from a pool of `3` winning tickets (`N_c`) to the total number of combinations of `4` out of `7` (N).
`P_c=C_3^2*C_4^2`

The numerator `N_c` equals to the number of combinations of `2` winning tickets out of `3` available `C_3^2=(3!)/(2!*1!)=3` multiplied by the number of combinations of `2` non-winning tickets out of `4` available `C_4^2=(4!)/(2!*2!)=6`.
So, numerator is
`N_c=C_3^2*C_4^2=3*6=18`

The denominator is
`N=C_7^4=(7!)/(4!*3!)=35`

So, the probability of event (c) is
`P_c=N_c/N=(3*6)/35=18/35`

Similarly, for case (d) we have
`N_d=C_3^3*C_4^1=1*4=4`
`P_d=N_d/N=4/35`

The total of probabilities of events (c) and (d) is
`P=P_c+P_d=18/35+4/35=22/35`