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Question

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A). $\dfrac{9}{{20}}$

B). $\dfrac{3}{{16}}$

C). $\dfrac{1}{6}$

D). $\dfrac{1}{9}$

Answer

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Given,

No. of Aces = 4

No. of Kings= 4

No. of Queens = 4

No. of Jacks= 4

So, the total number of cards we have = 16

Now, we need to select 2 cards out of these, which can be done in ${}^{16}{C_2}$ ways

Now, we need to find the probability of at least an ace.

And we know probability of an event = $\dfrac{{{\text{No}}{\text{. of favourable outcome}}}}{{{\text{Total no}}{\text{. of sample}}}} = \dfrac{{n(E)}}{{n(S)}}$

So , we can select at least one ace in two ways

Case1:- 1 Ace + 1 other card

1 Ace can be selected from 4 Aces in ${}^4{C_1}$ ways

1 other card from leftover 12 cards can be selected in ${}^{12}{C_1}$ways

So , selecting 1 Ace and 1 other card can be done in ${}^4{C_1} \times {}^{12}{C_1}$ways

Therefore, Probability of 1 Ace and 1 other card = $\dfrac{{{}^4{C_1} \times {}^{12}{C_1}}}{{{}^{16}{C_2}}}$

Case2:- both the cards are Ace

2 Aces can be selected from 4 aces in ${}^4{C_2}$ways

Probability of 2 Ace = $\dfrac{{{}^4{C_2}}}{{{}^{16}{C_2}}}$

Adding both the cases , we’ll get the probability of at least 2 aces

Therefore, the probability of at least 2 aces = $\dfrac{{{}^4{C_1} \times {}^{12}{C_1}}}{{{}^{16}{C_2}}}$+$\dfrac{{{}^4{C_2}}}{{{}^{16}{C_2}}}$

By using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ expanding all the terms.

$

= \dfrac{{4 \times 12 \times 2! \times 14!}}{{16 \times 15 \times 14!}} + \dfrac{{4 \times 3 \times 2! \times 14!}}{{2! \times 16 \times 15 \times 14!}} \\

= \dfrac{{4 \times 12 \times 2}}{{16 \times 15}} + \dfrac{{4 \times 3}}{{16 \times 15}} \\

= \dfrac{2}{5} + \dfrac{1}{{20}} \\

= \dfrac{9}{{20}} $