Part 2 of 3
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Let there be two players in a game, Player 1 and Player 2. Consider a jar containing 5 snakes. 3 of the snakes in the jar are venomous, while the remaining 2 are non-venomous. In the game, both the players have to put their hand in the jar one after the other and pick a snake out. Each snake, if picked out of the jar, will bite the player's hand. The event of picking a venomous snake, or equivalently, a venomous snake's bite will earn the player zero points. On the other hand, the event of picking a non-venomous snake, or equivalently, a non-venomous snake's bite will earn the player one point.
Let $X$ denote Player 1 's pick and let $Y$ denote Player 2 's pick. Suppose Player 1 is the first to pick out a snake.
The expected value of Player 1's pick is: $E(X)=0.4$.
(Express your answer as a fraction or round your answer to two decimal places.)
The expected value of Player 2's pick is: $E(Y)=$
(Express your answer as a fraction or round your answer to two decimal places.)

Step 1 ：Let's denote the event of Player 1 picking a venomous snake as V1 and picking a non-venomous snake as NV1. Similarly, for Player 2, let's denote the event of picking a venomous snake as V2 and picking a non-venomous snake as NV2.

Step 2 ：The probability of Player 1 picking a venomous snake, P(V1), is 3/5 = 0.6 and the probability of picking a non-venomous snake, P(NV1), is 2/5 = 0.4.

Step 3 ：If Player 1 picks a venomous snake, there are 2 venomous and 2 non-venomous snakes left. So, the probability of Player 2 picking a venomous snake after Player 1 has picked a venomous snake, P(V2|V1), is 2/4 = 0.5 and the probability of picking a non-venomous snake, P(NV2|V1), is also 2/4 = 0.5.

Step 4 ：If Player 1 picks a non-venomous snake, there are 3 venomous and 1 non-venomous snake left. So, the probability of Player 2 picking a venomous snake after Player 1 has picked a non-venomous snake, P(V2|NV1), is 3/4 = 0.75 and the probability of picking a non-venomous snake, P(NV2|NV1), is 1/4 = 0.25.

Step 5 ：The expected value of Player 2's pick after Player 1 has picked a venomous snake is E(Y|V1) = P(V2|V1)*0 + P(NV2|V1)*1 = 0.5*0 + 0.5*1 = 0.5.

Step 6 ：The expected value of Player 2's pick after Player 1 has picked a non-venomous snake is E(Y|NV1) = P(V2|NV1)*0 + P(NV2|NV1)*1 = 0.75*0 + 0.25*1 = 0.25.

Step 7 ：The total expected value of Player 2's pick is E(Y) = P(V1)*E(Y|V1) + P(NV1)*E(Y|NV1) = 0.6*0.5 + 0.4*0.25 = 0.4.

Step 8 ：Final Answer: The expected value of Player 2's pick is \(\boxed{0.4}\).