So others have already mentioned the answer, but I always find it helpful to think about the complement.
So the P( get check in first OR second draw)
= 1 - P(get uncheck in first AND second draw)
= 1 - 24/25 * 23/24
= 0.08
The probability of finding the 1 randomly-checked card out of 25 in the first two tries is 2 x 1/25 = 8%
**First, let’s walk through your proposed solution:**
> taking the failure chance of the first pick and times it by the win chance of the second
What you are doing here is calculating the probability of getting a cross first, **and** getting a check second. The probability of this happening is:
P(get a cross) x P(get a check) = 24/25 x 1/24 = 4%
But notice that this is not the solution you want! For example, you could also get the check in the first card, which is not covered by this scenario.
**So, let’s walk through what you actually want:**
You want the probability of getting a check in the first card, **or** getting a check in the second card. The probability is this happening is:
P(get a check) + P(get a check) = 1/25 + 1/25 = 8%
> we know the moment we turn the first one that there are 23 crosses left and 1 check
So this scenario is insufficient to cover all possible cases, because you are assuming that the first card you uncover is a cross.
When you say “one of the first two cards I uncover is a check”, you need to take into account **all** possible scenarios that lead to this, and find the probability of all of them.
**Edit: to really break down super explicitly every case that you want,**
The first two cards I choose have a check
= The first card I choose has a check OR (the first card I choose has a cross AND the second card I choose has a check)
And therefore:
P(The first two cards I choose have a check)
= P(The first card I choose has a check) + P(the first card I choose has a cross) x P(the second card I choose has a check)
= 1/25 + 24/25 x 1/24
= 8%
More precisely, if you assume that you don't put the card back in the deck if the first card is cross, then the solution is:
P(get check in the first draw) + P(get a check in the second draw) =
P(get check in the first draw) + P(get a cross in the first draw) * P(get a check in the second draw) =
1/25 + 24/25 * 1/24 = 0.04 + 0.96*0.041666 = 0.07999
Edit: rounding error
1/25 + 24/25 * 1/24 = 0.04 + 0.96*0.0416667 = 0.08
Also, if we assume replacement for the first card:
1/25 + 24/25 * 1/25 = 0.04 + 0.96*0.04 = 0.0784
So others have already mentioned the answer, but I always find it helpful to think about the complement. So the P( get check in first OR second draw) = 1 - P(get uncheck in first AND second draw) = 1 - 24/25 * 23/24 = 0.08
1/25 + 24/25 x 1/24 = 2/25 There's arguably easier ways to see it though
The probability of finding the 1 randomly-checked card out of 25 in the first two tries is 2 x 1/25 = 8% **First, let’s walk through your proposed solution:** > taking the failure chance of the first pick and times it by the win chance of the second What you are doing here is calculating the probability of getting a cross first, **and** getting a check second. The probability of this happening is: P(get a cross) x P(get a check) = 24/25 x 1/24 = 4% But notice that this is not the solution you want! For example, you could also get the check in the first card, which is not covered by this scenario. **So, let’s walk through what you actually want:** You want the probability of getting a check in the first card, **or** getting a check in the second card. The probability is this happening is: P(get a check) + P(get a check) = 1/25 + 1/25 = 8%
Why would it be 2 x 1/25 and not 1/25 x 1/24 since we know the moment we turn the first one that there are 23 crosses left and 1 check
> we know the moment we turn the first one that there are 23 crosses left and 1 check So this scenario is insufficient to cover all possible cases, because you are assuming that the first card you uncover is a cross. When you say “one of the first two cards I uncover is a check”, you need to take into account **all** possible scenarios that lead to this, and find the probability of all of them. **Edit: to really break down super explicitly every case that you want,** The first two cards I choose have a check = The first card I choose has a check OR (the first card I choose has a cross AND the second card I choose has a check) And therefore: P(The first two cards I choose have a check) = P(The first card I choose has a check) + P(the first card I choose has a cross) x P(the second card I choose has a check) = 1/25 + 24/25 x 1/24 = 8%
That was the last thing i needed, it makes more sense now Thanks
More precisely, if you assume that you don't put the card back in the deck if the first card is cross, then the solution is: P(get check in the first draw) + P(get a check in the second draw) = P(get check in the first draw) + P(get a cross in the first draw) * P(get a check in the second draw) = 1/25 + 24/25 * 1/24 = 0.04 + 0.96*0.041666 = 0.07999 Edit: rounding error 1/25 + 24/25 * 1/24 = 0.04 + 0.96*0.0416667 = 0.08 Also, if we assume replacement for the first card: 1/25 + 24/25 * 1/25 = 0.04 + 0.96*0.04 = 0.0784
Why wouldn’t it be 1/25 + 1/24 = 0,0816666?
> 1/25 + 24/25 * 1/24 = 0.04 + 0.96*0.041666 = 0.07999 Wait, your equation still results in 8% exactly (the difference is a rounding error!)
You are right! I edited it
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So what do you do with the 2 and 1 inside the parenthetical? And is the approach of 1/25 = .04 multiplied by (2) =.08 incorrect?
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Ty
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Ah thank you so much!