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A beg contains 4 white and 6 red balls. Four balls are drawn at random, then match the probability of the numbers of white balls drawn. Where P(E) denotes the probability of an event E.

Then Match List-I with List-II

LIST-ILIST-II
A.P (no white ball)I.821\frac{8}{21}
B.P (all white balls)II.114\frac{1}{14}
C.P (one white ball)III.614\frac{6}{14}
D.P (two white balls)IV.1210\frac{1}{210}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Total selections are (104)=210\binom{10}{4}=210. No white gives (64)=15\binom{6}{4}=15, so 114\frac{1}{14} (II). All white gives 1, so 1210\frac{1}{210} (IV). One white gives 4×20=804\times 20=80, so 821\frac{8}{21} (I). Two white gives 6×15=906\times 15=90, so 614\frac{6}{14} (III).

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